today’s outline - april 23, 2015segre/phys406/15s/lecture_23.pdftoday’s outline - april 23, 2015...
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Today’s Outline - April 23, 2015
• The Born Series
• The EPR paradox
• Problem 12.1
• Bell’s inequality
• The EPR experiment
Homework Assignment #10:Chapter 11:2,4,5,7,9,20due Tuesday, April 28, 2015
Final Exam: Monday, May 4, 201514:00-16:00, Herman Hall Ballroom
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 1 / 22
Today’s Outline - April 23, 2015
• The Born Series
• The EPR paradox
• Problem 12.1
• Bell’s inequality
• The EPR experiment
Homework Assignment #10:Chapter 11:2,4,5,7,9,20due Tuesday, April 28, 2015
Final Exam: Monday, May 4, 201514:00-16:00, Herman Hall Ballroom
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 1 / 22
Today’s Outline - April 23, 2015
• The Born Series
• The EPR paradox
• Problem 12.1
• Bell’s inequality
• The EPR experiment
Homework Assignment #10:Chapter 11:2,4,5,7,9,20due Tuesday, April 28, 2015
Final Exam: Monday, May 4, 201514:00-16:00, Herman Hall Ballroom
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 1 / 22
Today’s Outline - April 23, 2015
• The Born Series
• The EPR paradox
• Problem 12.1
• Bell’s inequality
• The EPR experiment
Homework Assignment #10:Chapter 11:2,4,5,7,9,20due Tuesday, April 28, 2015
Final Exam: Monday, May 4, 201514:00-16:00, Herman Hall Ballroom
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 1 / 22
Today’s Outline - April 23, 2015
• The Born Series
• The EPR paradox
• Problem 12.1
• Bell’s inequality
• The EPR experiment
Homework Assignment #10:Chapter 11:2,4,5,7,9,20due Tuesday, April 28, 2015
Final Exam: Monday, May 4, 201514:00-16:00, Herman Hall Ballroom
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 1 / 22
Today’s Outline - April 23, 2015
• The Born Series
• The EPR paradox
• Problem 12.1
• Bell’s inequality
• The EPR experiment
Homework Assignment #10:Chapter 11:2,4,5,7,9,20due Tuesday, April 28, 2015
Final Exam: Monday, May 4, 201514:00-16:00, Herman Hall Ballroom
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 1 / 22
Today’s Outline - April 23, 2015
• The Born Series
• The EPR paradox
• Problem 12.1
• Bell’s inequality
• The EPR experiment
Homework Assignment #10:Chapter 11:2,4,5,7,9,20due Tuesday, April 28, 2015
Final Exam: Monday, May 4, 201514:00-16:00, Herman Hall Ballroom
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 1 / 22
Today’s Outline - April 23, 2015
• The Born Series
• The EPR paradox
• Problem 12.1
• Bell’s inequality
• The EPR experiment
Homework Assignment #10:Chapter 11:2,4,5,7,9,20due Tuesday, April 28, 2015
Final Exam: Monday, May 4, 201514:00-16:00, Herman Hall Ballroom
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 1 / 22
The impulse approximation
θF
b
In classical physics, the impulse approximation assumes that the particlekeeps going in a straight line and a transverse impulse is computed to givethe final scattering angle
I =
∫F⊥ dt −→ θ ∼= tan−1
(I
p
)this first-order correction is analgous to the first Born approximation wherethe incident particle is unaffected
following this analogy, we can generate a series of scattering correctionscalled the Born series
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 2 / 22
The impulse approximation
θF
b
In classical physics, the impulse approximation assumes that the particlekeeps going in a straight line and a transverse impulse is computed to givethe final scattering angle
I =
∫F⊥ dt −→ θ ∼= tan−1
(I
p
)this first-order correction is analgous to the first Born approximation wherethe incident particle is unaffected
following this analogy, we can generate a series of scattering correctionscalled the Born series
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 2 / 22
The impulse approximation
θF
b
In classical physics, the impulse approximation assumes that the particlekeeps going in a straight line and a transverse impulse is computed to givethe final scattering angle
I =
∫F⊥ dt
−→ θ ∼= tan−1(I
p
)this first-order correction is analgous to the first Born approximation wherethe incident particle is unaffected
following this analogy, we can generate a series of scattering correctionscalled the Born series
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 2 / 22
The impulse approximation
θF
b
In classical physics, the impulse approximation assumes that the particlekeeps going in a straight line and a transverse impulse is computed to givethe final scattering angle
I =
∫F⊥ dt −→ θ ∼= tan−1
(I
p
)
this first-order correction is analgous to the first Born approximation wherethe incident particle is unaffected
following this analogy, we can generate a series of scattering correctionscalled the Born series
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 2 / 22
The impulse approximation
θF
b
In classical physics, the impulse approximation assumes that the particlekeeps going in a straight line and a transverse impulse is computed to givethe final scattering angle
I =
∫F⊥ dt −→ θ ∼= tan−1
(I
p
)this first-order correction is analgous to the first Born approximation wherethe incident particle is unaffected
following this analogy, we can generate a series of scattering correctionscalled the Born series
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 2 / 22
The impulse approximation
θF
b
In classical physics, the impulse approximation assumes that the particlekeeps going in a straight line and a transverse impulse is computed to givethe final scattering angle
I =
∫F⊥ dt −→ θ ∼= tan−1
(I
p
)this first-order correction is analgous to the first Born approximation wherethe incident particle is unaffected
following this analogy, we can generate a series of scattering correctionscalled the Born series
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 2 / 22
The Born series
Starting with the integral Schrodinger equation
ψ(r) = ψ0(r) +
∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m
2π~2e ikr
rthis can be written as
now replace ψ on the rightside by it’s integral represen-tation
ψ = ψ0 +
∫gVψ
ψ = ψ0 +
∫gVψ0 +
∫ ∫gVgVψ
repeating the process, we have a recursive integral equation
ψ = ψ0 +
∫gVψ0 +
∫ ∫gVgVψ0 +
∫ ∫ ∫gVgVgVψ0 + · · ·
+V
g + + + . . .ψ
0ψ
0
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 3 / 22
The Born series
Starting with the integral Schrodinger equation
ψ(r) = ψ0(r) +
∫g(r − r0)V (r0)ψ(r0) dr0,
g(r) = − m
2π~2e ikr
rthis can be written as
now replace ψ on the rightside by it’s integral represen-tation
ψ = ψ0 +
∫gVψ
ψ = ψ0 +
∫gVψ0 +
∫ ∫gVgVψ
repeating the process, we have a recursive integral equation
ψ = ψ0 +
∫gVψ0 +
∫ ∫gVgVψ0 +
∫ ∫ ∫gVgVgVψ0 + · · ·
+V
g + + + . . .ψ
0ψ
0
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 3 / 22
The Born series
Starting with the integral Schrodinger equation
ψ(r) = ψ0(r) +
∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m
2π~2e ikr
r
this can be written as
now replace ψ on the rightside by it’s integral represen-tation
ψ = ψ0 +
∫gVψ
ψ = ψ0 +
∫gVψ0 +
∫ ∫gVgVψ
repeating the process, we have a recursive integral equation
ψ = ψ0 +
∫gVψ0 +
∫ ∫gVgVψ0 +
∫ ∫ ∫gVgVgVψ0 + · · ·
+V
g + + + . . .ψ
0ψ
0
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 3 / 22
The Born series
Starting with the integral Schrodinger equation
ψ(r) = ψ0(r) +
∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m
2π~2e ikr
rthis can be written as
now replace ψ on the rightside by it’s integral represen-tation
ψ = ψ0 +
∫gVψ
ψ = ψ0 +
∫gVψ0 +
∫ ∫gVgVψ
repeating the process, we have a recursive integral equation
ψ = ψ0 +
∫gVψ0 +
∫ ∫gVgVψ0 +
∫ ∫ ∫gVgVgVψ0 + · · ·
+V
g + + + . . .ψ
0ψ
0
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 3 / 22
The Born series
Starting with the integral Schrodinger equation
ψ(r) = ψ0(r) +
∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m
2π~2e ikr
rthis can be written as
now replace ψ on the rightside by it’s integral represen-tation
ψ = ψ0 +
∫gVψ
ψ = ψ0 +
∫gVψ0 +
∫ ∫gVgVψ
repeating the process, we have a recursive integral equation
ψ = ψ0 +
∫gVψ0 +
∫ ∫gVgVψ0 +
∫ ∫ ∫gVgVgVψ0 + · · ·
+V
g + + + . . .ψ
0ψ
0
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 3 / 22
The Born series
Starting with the integral Schrodinger equation
ψ(r) = ψ0(r) +
∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m
2π~2e ikr
rthis can be written as
now replace ψ on the rightside by it’s integral represen-tation
ψ = ψ0 +
∫gVψ
ψ = ψ0 +
∫gVψ0 +
∫ ∫gVgVψ
repeating the process, we have a recursive integral equation
ψ = ψ0 +
∫gVψ0 +
∫ ∫gVgVψ0 +
∫ ∫ ∫gVgVgVψ0 + · · ·
+V
g + + + . . .ψ
0ψ
0
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 3 / 22
The Born series
Starting with the integral Schrodinger equation
ψ(r) = ψ0(r) +
∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m
2π~2e ikr
rthis can be written as
now replace ψ on the rightside by it’s integral represen-tation
ψ = ψ0 +
∫gVψ
ψ = ψ0 +
∫gVψ0 +
∫ ∫gVgVψ
repeating the process, we have a recursive integral equation
ψ = ψ0 +
∫gVψ0 +
∫ ∫gVgVψ0 +
∫ ∫ ∫gVgVgVψ0 + · · ·
+V
g + + + . . .ψ
0ψ
0
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 3 / 22
The Born series
Starting with the integral Schrodinger equation
ψ(r) = ψ0(r) +
∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m
2π~2e ikr
rthis can be written as
now replace ψ on the rightside by it’s integral represen-tation
ψ = ψ0 +
∫gVψ
ψ = ψ0 +
∫gVψ0 +
∫ ∫gVgVψ
repeating the process, we have a recursive integral equation
ψ = ψ0 +
∫gVψ0 +
∫ ∫gVgVψ0 +
∫ ∫ ∫gVgVgVψ0 + · · ·
+V
g + + + . . .ψ
0ψ
0
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 3 / 22
The Born series
Starting with the integral Schrodinger equation
ψ(r) = ψ0(r) +
∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m
2π~2e ikr
rthis can be written as
now replace ψ on the rightside by it’s integral represen-tation
ψ = ψ0 +
∫gVψ
ψ = ψ0 +
∫gVψ0 +
∫ ∫gVgVψ
repeating the process, we have a recursive integral equation
ψ = ψ0 +
∫gVψ0 +
∫ ∫gVgVψ0 +
∫ ∫ ∫gVgVgVψ0 + · · ·
+V
g + + + . . .ψ
0ψ
0
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 3 / 22
The Born series
Starting with the integral Schrodinger equation
ψ(r) = ψ0(r) +
∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m
2π~2e ikr
rthis can be written as
now replace ψ on the rightside by it’s integral represen-tation
ψ = ψ0 +
∫gVψ
ψ = ψ0 +
∫gVψ0 +
∫ ∫gVgVψ
repeating the process, we have a recursive integral equation
ψ = ψ0 +
∫gVψ0 +
∫ ∫gVgVψ0 +
∫ ∫ ∫gVgVgVψ0 + · · ·
+V
g + + + . . .ψ
0ψ
0
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 3 / 22
Quantum paradoxes and other fun stuff
“About your cat, Mr. Schrodinger – I have good news and bad news . . . ”C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 4 / 22
Einstein Podolsky Rosen paradox
“Can quantum-mechanical description of physical reality be consideredcomplete?,” Physical Review 47, 777-779 (1935)
“If, without in any way disturbing a system, we can predictwith certainty (i.e., with probability equal to unity) thevalue of a physical quantity, then there exists an elementof physical reality corresponding to this physical quantity.”
If ψ is an eigenfuction of an opera-tor A, then
for example take momentum as theoperator
Aψ = aψ
ψ = e ip0x/~
p =~i
∂
∂x
pψ =~i
∂
∂xe ip0x/~ =
~i
i
~p0e
ip0x/~ = p0ψ
thus the momentum in state ψ is said to be real
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 5 / 22
Einstein Podolsky Rosen paradox
“Can quantum-mechanical description of physical reality be consideredcomplete?,” Physical Review 47, 777-779 (1935)
“If, without in any way disturbing a system, we can predictwith certainty (i.e., with probability equal to unity) thevalue of a physical quantity, then there exists an elementof physical reality corresponding to this physical quantity.”
If ψ is an eigenfuction of an opera-tor A, then
for example take momentum as theoperator
Aψ = aψ
ψ = e ip0x/~
p =~i
∂
∂x
pψ =~i
∂
∂xe ip0x/~ =
~i
i
~p0e
ip0x/~ = p0ψ
thus the momentum in state ψ is said to be real
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 5 / 22
Einstein Podolsky Rosen paradox
“Can quantum-mechanical description of physical reality be consideredcomplete?,” Physical Review 47, 777-779 (1935)
“If, without in any way disturbing a system, we can predictwith certainty (i.e., with probability equal to unity) thevalue of a physical quantity, then there exists an elementof physical reality corresponding to this physical quantity.”
If ψ is an eigenfuction of an opera-tor A, then
for example take momentum as theoperator
Aψ = aψ
ψ = e ip0x/~
p =~i
∂
∂x
pψ =~i
∂
∂xe ip0x/~ =
~i
i
~p0e
ip0x/~ = p0ψ
thus the momentum in state ψ is said to be real
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 5 / 22
Einstein Podolsky Rosen paradox
“Can quantum-mechanical description of physical reality be consideredcomplete?,” Physical Review 47, 777-779 (1935)
“If, without in any way disturbing a system, we can predictwith certainty (i.e., with probability equal to unity) thevalue of a physical quantity, then there exists an elementof physical reality corresponding to this physical quantity.”
If ψ is an eigenfuction of an opera-tor A, then
for example take momentum as theoperator
Aψ = aψ
ψ = e ip0x/~
p =~i
∂
∂x
pψ =~i
∂
∂xe ip0x/~ =
~i
i
~p0e
ip0x/~ = p0ψ
thus the momentum in state ψ is said to be real
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 5 / 22
Einstein Podolsky Rosen paradox
“Can quantum-mechanical description of physical reality be consideredcomplete?,” Physical Review 47, 777-779 (1935)
“If, without in any way disturbing a system, we can predictwith certainty (i.e., with probability equal to unity) thevalue of a physical quantity, then there exists an elementof physical reality corresponding to this physical quantity.”
If ψ is an eigenfuction of an opera-tor A, then
for example take momentum as theoperator
Aψ = aψ
ψ = e ip0x/~
p =~i
∂
∂x
pψ =~i
∂
∂xe ip0x/~ =
~i
i
~p0e
ip0x/~ = p0ψ
thus the momentum in state ψ is said to be real
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 5 / 22
Einstein Podolsky Rosen paradox
“Can quantum-mechanical description of physical reality be consideredcomplete?,” Physical Review 47, 777-779 (1935)
“If, without in any way disturbing a system, we can predictwith certainty (i.e., with probability equal to unity) thevalue of a physical quantity, then there exists an elementof physical reality corresponding to this physical quantity.”
If ψ is an eigenfuction of an opera-tor A, then
for example take momentum as theoperator
Aψ = aψ
ψ = e ip0x/~
p =~i
∂
∂x
pψ =~i
∂
∂xe ip0x/~ =
~i
i
~p0e
ip0x/~ = p0ψ
thus the momentum in state ψ is said to be real
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 5 / 22
Einstein Podolsky Rosen paradox
“Can quantum-mechanical description of physical reality be consideredcomplete?,” Physical Review 47, 777-779 (1935)
“If, without in any way disturbing a system, we can predictwith certainty (i.e., with probability equal to unity) thevalue of a physical quantity, then there exists an elementof physical reality corresponding to this physical quantity.”
If ψ is an eigenfuction of an opera-tor A, then
for example take momentum as theoperator
Aψ = aψ
ψ = e ip0x/~
p =~i
∂
∂x
pψ =~i
∂
∂xe ip0x/~ =
~i
i
~p0e
ip0x/~ = p0ψ
thus the momentum in state ψ is said to be real
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 5 / 22
Einstein Podolsky Rosen paradox
“Can quantum-mechanical description of physical reality be consideredcomplete?,” Physical Review 47, 777-779 (1935)
“If, without in any way disturbing a system, we can predictwith certainty (i.e., with probability equal to unity) thevalue of a physical quantity, then there exists an elementof physical reality corresponding to this physical quantity.”
If ψ is an eigenfuction of an opera-tor A, then
for example take momentum as theoperator
Aψ = aψ
ψ = e ip0x/~
p =~i
∂
∂x
pψ =~i
∂
∂xe ip0x/~
=~i
i
~p0e
ip0x/~ = p0ψ
thus the momentum in state ψ is said to be real
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 5 / 22
Einstein Podolsky Rosen paradox
“Can quantum-mechanical description of physical reality be consideredcomplete?,” Physical Review 47, 777-779 (1935)
“If, without in any way disturbing a system, we can predictwith certainty (i.e., with probability equal to unity) thevalue of a physical quantity, then there exists an elementof physical reality corresponding to this physical quantity.”
If ψ is an eigenfuction of an opera-tor A, then
for example take momentum as theoperator
Aψ = aψ
ψ = e ip0x/~
p =~i
∂
∂x
pψ =~i
∂
∂xe ip0x/~ =
~i
i
~p0e
ip0x/~
= p0ψ
thus the momentum in state ψ is said to be real
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 5 / 22
Einstein Podolsky Rosen paradox
“Can quantum-mechanical description of physical reality be consideredcomplete?,” Physical Review 47, 777-779 (1935)
“If, without in any way disturbing a system, we can predictwith certainty (i.e., with probability equal to unity) thevalue of a physical quantity, then there exists an elementof physical reality corresponding to this physical quantity.”
If ψ is an eigenfuction of an opera-tor A, then
for example take momentum as theoperator
Aψ = aψ
ψ = e ip0x/~
p =~i
∂
∂x
pψ =~i
∂
∂xe ip0x/~ =
~i
i
~p0e
ip0x/~ = p0ψ
thus the momentum in state ψ is said to be real
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 5 / 22
Einstein Podolsky Rosen paradox
“Can quantum-mechanical description of physical reality be consideredcomplete?,” Physical Review 47, 777-779 (1935)
“If, without in any way disturbing a system, we can predictwith certainty (i.e., with probability equal to unity) thevalue of a physical quantity, then there exists an elementof physical reality corresponding to this physical quantity.”
If ψ is an eigenfuction of an opera-tor A, then
for example take momentum as theoperator
Aψ = aψ
ψ = e ip0x/~
p =~i
∂
∂x
pψ =~i
∂
∂xe ip0x/~ =
~i
i
~p0e
ip0x/~ = p0ψ
thus the momentum in state ψ is said to be real
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 5 / 22
Einstein Podolsky Rosen paradox
If Aψ = aψ does not hold, how-ever, A cannot be said to have aparticular value as we know fromthe position operator q
qψ = xe ip0x/~ 6= aψ
P(a, b) =
∫ b
aψ∗ψ dx =
∫ b
adx = b − a
there is an equal probability of measuring any value of the position
this is a specific instance of the fact that if two operators do not commute[A,B] 6= 0 then precise knowledge of one precludes such knowledge of theother
the authors thus conclude that
1 the quantum-mechanical description of reality given by the wavefunction is not complete, or
2 when the operators corresponding to two physical quantities do notcommute the two quantities cannot have simultaneous reality
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 6 / 22
Einstein Podolsky Rosen paradox
If Aψ = aψ does not hold, how-ever, A cannot be said to have aparticular value as we know fromthe position operator q
qψ = xe ip0x/~
6= aψ
P(a, b) =
∫ b
aψ∗ψ dx =
∫ b
adx = b − a
there is an equal probability of measuring any value of the position
this is a specific instance of the fact that if two operators do not commute[A,B] 6= 0 then precise knowledge of one precludes such knowledge of theother
the authors thus conclude that
1 the quantum-mechanical description of reality given by the wavefunction is not complete, or
2 when the operators corresponding to two physical quantities do notcommute the two quantities cannot have simultaneous reality
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 6 / 22
Einstein Podolsky Rosen paradox
If Aψ = aψ does not hold, how-ever, A cannot be said to have aparticular value as we know fromthe position operator q
qψ = xe ip0x/~ 6= aψ
P(a, b) =
∫ b
aψ∗ψ dx =
∫ b
adx = b − a
there is an equal probability of measuring any value of the position
this is a specific instance of the fact that if two operators do not commute[A,B] 6= 0 then precise knowledge of one precludes such knowledge of theother
the authors thus conclude that
1 the quantum-mechanical description of reality given by the wavefunction is not complete, or
2 when the operators corresponding to two physical quantities do notcommute the two quantities cannot have simultaneous reality
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 6 / 22
Einstein Podolsky Rosen paradox
If Aψ = aψ does not hold, how-ever, A cannot be said to have aparticular value as we know fromthe position operator q
qψ = xe ip0x/~ 6= aψ
P(a, b) =
∫ b
aψ∗ψ dx
=
∫ b
adx = b − a
there is an equal probability of measuring any value of the position
this is a specific instance of the fact that if two operators do not commute[A,B] 6= 0 then precise knowledge of one precludes such knowledge of theother
the authors thus conclude that
1 the quantum-mechanical description of reality given by the wavefunction is not complete, or
2 when the operators corresponding to two physical quantities do notcommute the two quantities cannot have simultaneous reality
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 6 / 22
Einstein Podolsky Rosen paradox
If Aψ = aψ does not hold, how-ever, A cannot be said to have aparticular value as we know fromthe position operator q
qψ = xe ip0x/~ 6= aψ
P(a, b) =
∫ b
aψ∗ψ dx =
∫ b
adx
= b − a
there is an equal probability of measuring any value of the position
this is a specific instance of the fact that if two operators do not commute[A,B] 6= 0 then precise knowledge of one precludes such knowledge of theother
the authors thus conclude that
1 the quantum-mechanical description of reality given by the wavefunction is not complete, or
2 when the operators corresponding to two physical quantities do notcommute the two quantities cannot have simultaneous reality
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 6 / 22
Einstein Podolsky Rosen paradox
If Aψ = aψ does not hold, how-ever, A cannot be said to have aparticular value as we know fromthe position operator q
qψ = xe ip0x/~ 6= aψ
P(a, b) =
∫ b
aψ∗ψ dx =
∫ b
adx = b − a
there is an equal probability of measuring any value of the position
this is a specific instance of the fact that if two operators do not commute[A,B] 6= 0 then precise knowledge of one precludes such knowledge of theother
the authors thus conclude that
1 the quantum-mechanical description of reality given by the wavefunction is not complete, or
2 when the operators corresponding to two physical quantities do notcommute the two quantities cannot have simultaneous reality
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 6 / 22
Einstein Podolsky Rosen paradox
If Aψ = aψ does not hold, how-ever, A cannot be said to have aparticular value as we know fromthe position operator q
qψ = xe ip0x/~ 6= aψ
P(a, b) =
∫ b
aψ∗ψ dx =
∫ b
adx = b − a
there is an equal probability of measuring any value of the position
this is a specific instance of the fact that if two operators do not commute[A,B] 6= 0 then precise knowledge of one precludes such knowledge of theother
the authors thus conclude that
1 the quantum-mechanical description of reality given by the wavefunction is not complete, or
2 when the operators corresponding to two physical quantities do notcommute the two quantities cannot have simultaneous reality
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 6 / 22
Einstein Podolsky Rosen paradox
If Aψ = aψ does not hold, how-ever, A cannot be said to have aparticular value as we know fromthe position operator q
qψ = xe ip0x/~ 6= aψ
P(a, b) =
∫ b
aψ∗ψ dx =
∫ b
adx = b − a
there is an equal probability of measuring any value of the position
this is a specific instance of the fact that if two operators do not commute[A,B] 6= 0 then precise knowledge of one precludes such knowledge of theother
the authors thus conclude that
1 the quantum-mechanical description of reality given by the wavefunction is not complete, or
2 when the operators corresponding to two physical quantities do notcommute the two quantities cannot have simultaneous reality
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 6 / 22
Einstein Podolsky Rosen paradox
If Aψ = aψ does not hold, how-ever, A cannot be said to have aparticular value as we know fromthe position operator q
qψ = xe ip0x/~ 6= aψ
P(a, b) =
∫ b
aψ∗ψ dx =
∫ b
adx = b − a
there is an equal probability of measuring any value of the position
this is a specific instance of the fact that if two operators do not commute[A,B] 6= 0 then precise knowledge of one precludes such knowledge of theother
the authors thus conclude that
1 the quantum-mechanical description of reality given by the wavefunction is not complete, or
2 when the operators corresponding to two physical quantities do notcommute the two quantities cannot have simultaneous reality
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 6 / 22
Einstein Podolsky Rosen paradox
If Aψ = aψ does not hold, how-ever, A cannot be said to have aparticular value as we know fromthe position operator q
qψ = xe ip0x/~ 6= aψ
P(a, b) =
∫ b
aψ∗ψ dx =
∫ b
adx = b − a
there is an equal probability of measuring any value of the position
this is a specific instance of the fact that if two operators do not commute[A,B] 6= 0 then precise knowledge of one precludes such knowledge of theother
the authors thus conclude that
1 the quantum-mechanical description of reality given by the wavefunction is not complete, or
2 when the operators corresponding to two physical quantities do notcommute the two quantities cannot have simultaneous reality
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 6 / 22
Einstein Podolsky Rosen paradox
If Aψ = aψ does not hold, how-ever, A cannot be said to have aparticular value as we know fromthe position operator q
qψ = xe ip0x/~ 6= aψ
P(a, b) =
∫ b
aψ∗ψ dx =
∫ b
adx = b − a
there is an equal probability of measuring any value of the position
this is a specific instance of the fact that if two operators do not commute[A,B] 6= 0 then precise knowledge of one precludes such knowledge of theother
the authors thus conclude that
1 the quantum-mechanical description of reality given by the wavefunction is not complete, or
2 when the operators corresponding to two physical quantities do notcommute the two quantities cannot have simultaneous reality
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 6 / 22
Einstein Podolsky Rosen paradox
Suppose that we have two systems, I and II which interact over a timeperiod 0 ≤ t ≤ T
if we know the state of each system before t = 0, then by using theSchrodinger equation, we can calculate the state of the combined systemI+II at any subsequent time, specifically t > T and we call it Ψ
suppose that a physical quantity Ahas eigenvalues and eigenfunctions
then the system-wide wavefunctionas a function of x1 is
where here ψn(x2) are the “coeffi-cients” of the expansion in un(x1)
a1, a2, a3, . . .
u1(x1), u2(x1), u3(x1), . . .
Ψ(x1, x2) =∞∑n=1
ψn(x2)un(x1)
if A is now measured and is found to have value ak then the first systemmust be left in state uk(x1) and the second system must be, thereforefound in state ψk(x2) so that Ψ = ψk(x2)uk(x1)
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 7 / 22
Einstein Podolsky Rosen paradox
Suppose that we have two systems, I and II which interact over a timeperiod 0 ≤ t ≤ T
if we know the state of each system before t = 0, then by using theSchrodinger equation, we can calculate the state of the combined systemI+II at any subsequent time, specifically t > T and we call it Ψ
suppose that a physical quantity Ahas eigenvalues and eigenfunctions
then the system-wide wavefunctionas a function of x1 is
where here ψn(x2) are the “coeffi-cients” of the expansion in un(x1)
a1, a2, a3, . . .
u1(x1), u2(x1), u3(x1), . . .
Ψ(x1, x2) =∞∑n=1
ψn(x2)un(x1)
if A is now measured and is found to have value ak then the first systemmust be left in state uk(x1) and the second system must be, thereforefound in state ψk(x2) so that Ψ = ψk(x2)uk(x1)
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 7 / 22
Einstein Podolsky Rosen paradox
Suppose that we have two systems, I and II which interact over a timeperiod 0 ≤ t ≤ T
if we know the state of each system before t = 0, then by using theSchrodinger equation, we can calculate the state of the combined systemI+II at any subsequent time, specifically t > T and we call it Ψ
suppose that a physical quantity Ahas eigenvalues
and eigenfunctions
then the system-wide wavefunctionas a function of x1 is
where here ψn(x2) are the “coeffi-cients” of the expansion in un(x1)
a1, a2, a3, . . .
u1(x1), u2(x1), u3(x1), . . .
Ψ(x1, x2) =∞∑n=1
ψn(x2)un(x1)
if A is now measured and is found to have value ak then the first systemmust be left in state uk(x1) and the second system must be, thereforefound in state ψk(x2) so that Ψ = ψk(x2)uk(x1)
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 7 / 22
Einstein Podolsky Rosen paradox
Suppose that we have two systems, I and II which interact over a timeperiod 0 ≤ t ≤ T
if we know the state of each system before t = 0, then by using theSchrodinger equation, we can calculate the state of the combined systemI+II at any subsequent time, specifically t > T and we call it Ψ
suppose that a physical quantity Ahas eigenvalues
and eigenfunctions
then the system-wide wavefunctionas a function of x1 is
where here ψn(x2) are the “coeffi-cients” of the expansion in un(x1)
a1, a2, a3, . . .
u1(x1), u2(x1), u3(x1), . . .
Ψ(x1, x2) =∞∑n=1
ψn(x2)un(x1)
if A is now measured and is found to have value ak then the first systemmust be left in state uk(x1) and the second system must be, thereforefound in state ψk(x2) so that Ψ = ψk(x2)uk(x1)
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 7 / 22
Einstein Podolsky Rosen paradox
Suppose that we have two systems, I and II which interact over a timeperiod 0 ≤ t ≤ T
if we know the state of each system before t = 0, then by using theSchrodinger equation, we can calculate the state of the combined systemI+II at any subsequent time, specifically t > T and we call it Ψ
suppose that a physical quantity Ahas eigenvalues and eigenfunctions
then the system-wide wavefunctionas a function of x1 is
where here ψn(x2) are the “coeffi-cients” of the expansion in un(x1)
a1, a2, a3, . . .
u1(x1), u2(x1), u3(x1), . . .
Ψ(x1, x2) =∞∑n=1
ψn(x2)un(x1)
if A is now measured and is found to have value ak then the first systemmust be left in state uk(x1) and the second system must be, thereforefound in state ψk(x2) so that Ψ = ψk(x2)uk(x1)
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 7 / 22
Einstein Podolsky Rosen paradox
Suppose that we have two systems, I and II which interact over a timeperiod 0 ≤ t ≤ T
if we know the state of each system before t = 0, then by using theSchrodinger equation, we can calculate the state of the combined systemI+II at any subsequent time, specifically t > T and we call it Ψ
suppose that a physical quantity Ahas eigenvalues and eigenfunctions
then the system-wide wavefunctionas a function of x1 is
where here ψn(x2) are the “coeffi-cients” of the expansion in un(x1)
a1, a2, a3, . . .
u1(x1), u2(x1), u3(x1), . . .
Ψ(x1, x2) =∞∑n=1
ψn(x2)un(x1)
if A is now measured and is found to have value ak then the first systemmust be left in state uk(x1) and the second system must be, thereforefound in state ψk(x2) so that Ψ = ψk(x2)uk(x1)
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 7 / 22
Einstein Podolsky Rosen paradox
Suppose that we have two systems, I and II which interact over a timeperiod 0 ≤ t ≤ T
if we know the state of each system before t = 0, then by using theSchrodinger equation, we can calculate the state of the combined systemI+II at any subsequent time, specifically t > T and we call it Ψ
suppose that a physical quantity Ahas eigenvalues and eigenfunctions
then the system-wide wavefunctionas a function of x1 is
where here ψn(x2) are the “coeffi-cients” of the expansion in un(x1)
a1, a2, a3, . . .
u1(x1), u2(x1), u3(x1), . . .
Ψ(x1, x2) =∞∑n=1
ψn(x2)un(x1)
if A is now measured and is found to have value ak then the first systemmust be left in state uk(x1) and the second system must be, thereforefound in state ψk(x2) so that Ψ = ψk(x2)uk(x1)
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 7 / 22
Einstein Podolsky Rosen paradox
Suppose that we have two systems, I and II which interact over a timeperiod 0 ≤ t ≤ T
if we know the state of each system before t = 0, then by using theSchrodinger equation, we can calculate the state of the combined systemI+II at any subsequent time, specifically t > T and we call it Ψ
suppose that a physical quantity Ahas eigenvalues and eigenfunctions
then the system-wide wavefunctionas a function of x1 is
where here ψn(x2) are the “coeffi-cients” of the expansion in un(x1)
a1, a2, a3, . . .
u1(x1), u2(x1), u3(x1), . . .
Ψ(x1, x2) =∞∑n=1
ψn(x2)un(x1)
if A is now measured and is found to have value ak then the first systemmust be left in state uk(x1) and the second system must be, thereforefound in state ψk(x2) so that Ψ = ψk(x2)uk(x1)
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 7 / 22
Einstein Podolsky Rosen paradox
Suppose that we have two systems, I and II which interact over a timeperiod 0 ≤ t ≤ T
if we know the state of each system before t = 0, then by using theSchrodinger equation, we can calculate the state of the combined systemI+II at any subsequent time, specifically t > T and we call it Ψ
suppose that a physical quantity Ahas eigenvalues and eigenfunctions
then the system-wide wavefunctionas a function of x1 is
where here ψn(x2) are the “coeffi-cients” of the expansion in un(x1)
a1, a2, a3, . . .
u1(x1), u2(x1), u3(x1), . . .
Ψ(x1, x2) =∞∑n=1
ψn(x2)un(x1)
if A is now measured and is found to have value ak then the first systemmust be left in state uk(x1) and the second system must be, thereforefound in state ψk(x2) so that Ψ = ψk(x2)uk(x1)
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 7 / 22
Einstein Podolsky Rosen paradox
Suppose that we have two systems, I and II which interact over a timeperiod 0 ≤ t ≤ T
if we know the state of each system before t = 0, then by using theSchrodinger equation, we can calculate the state of the combined systemI+II at any subsequent time, specifically t > T and we call it Ψ
suppose that a physical quantity Ahas eigenvalues and eigenfunctions
then the system-wide wavefunctionas a function of x1 is
where here ψn(x2) are the “coeffi-cients” of the expansion in un(x1)
a1, a2, a3, . . .
u1(x1), u2(x1), u3(x1), . . .
Ψ(x1, x2) =∞∑n=1
ψn(x2)un(x1)
if A is now measured and is found to have value ak then the first systemmust be left in state uk(x1) and the second system must be, thereforefound in state ψk(x2)
so that Ψ = ψk(x2)uk(x1)
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 7 / 22
Einstein Podolsky Rosen paradox
Suppose that we have two systems, I and II which interact over a timeperiod 0 ≤ t ≤ T
if we know the state of each system before t = 0, then by using theSchrodinger equation, we can calculate the state of the combined systemI+II at any subsequent time, specifically t > T and we call it Ψ
suppose that a physical quantity Ahas eigenvalues and eigenfunctions
then the system-wide wavefunctionas a function of x1 is
where here ψn(x2) are the “coeffi-cients” of the expansion in un(x1)
a1, a2, a3, . . .
u1(x1), u2(x1), u3(x1), . . .
Ψ(x1, x2) =∞∑n=1
ψn(x2)un(x1)
if A is now measured and is found to have value ak then the first systemmust be left in state uk(x1) and the second system must be, thereforefound in state ψk(x2) so that Ψ = ψk(x2)uk(x1)
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 7 / 22
Einstein Podolsky Rosen paradox
If instead of operator A, we chooseto expand the overall wavefunctionin terms of the eigenfunctions of B,then we have
b1, b2, b3, . . .
v1(x1), v2(x1), v3(x1), . . .
Ψ(x1, x2) =∞∑s=1
ϕs(x2)vs(x1)
if B is measured and found to be br , then the system can be said to be inthe state Ψ = ϕr (x2)vr (x1) and the second system must be in state ϕr (x2)
Thus, as a consequence of two different measurements made on System I,System II can be left in states with two different wave functions, when it isfar away from, and not interacting with System I
Thus one can assign two different wave functions, ψk(x2) and ϕr (x2), tothe same reality (System II after interaction with System I)
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 8 / 22
Einstein Podolsky Rosen paradox
If instead of operator A, we chooseto expand the overall wavefunctionin terms of the eigenfunctions of B,then we have
b1, b2, b3, . . .
v1(x1), v2(x1), v3(x1), . . .
Ψ(x1, x2) =∞∑s=1
ϕs(x2)vs(x1)
if B is measured and found to be br , then the system can be said to be inthe state Ψ = ϕr (x2)vr (x1) and the second system must be in state ϕr (x2)
Thus, as a consequence of two different measurements made on System I,System II can be left in states with two different wave functions, when it isfar away from, and not interacting with System I
Thus one can assign two different wave functions, ψk(x2) and ϕr (x2), tothe same reality (System II after interaction with System I)
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 8 / 22
Einstein Podolsky Rosen paradox
If instead of operator A, we chooseto expand the overall wavefunctionin terms of the eigenfunctions of B,then we have
b1, b2, b3, . . .
v1(x1), v2(x1), v3(x1), . . .
Ψ(x1, x2) =∞∑s=1
ϕs(x2)vs(x1)
if B is measured and found to be br , then the system can be said to be inthe state Ψ = ϕr (x2)vr (x1) and the second system must be in state ϕr (x2)
Thus, as a consequence of two different measurements made on System I,System II can be left in states with two different wave functions, when it isfar away from, and not interacting with System I
Thus one can assign two different wave functions, ψk(x2) and ϕr (x2), tothe same reality (System II after interaction with System I)
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 8 / 22
Einstein Podolsky Rosen paradox
If instead of operator A, we chooseto expand the overall wavefunctionin terms of the eigenfunctions of B,then we have
b1, b2, b3, . . .
v1(x1), v2(x1), v3(x1), . . .
Ψ(x1, x2) =∞∑s=1
ϕs(x2)vs(x1)
if B is measured and found to be br , then the system can be said to be inthe state Ψ = ϕr (x2)vr (x1) and the second system must be in state ϕr (x2)
Thus, as a consequence of two different measurements made on System I,System II can be left in states with two different wave functions, when it isfar away from, and not interacting with System I
Thus one can assign two different wave functions, ψk(x2) and ϕr (x2), tothe same reality (System II after interaction with System I)
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 8 / 22
Einstein Podolsky Rosen paradox
If instead of operator A, we chooseto expand the overall wavefunctionin terms of the eigenfunctions of B,then we have
b1, b2, b3, . . .
v1(x1), v2(x1), v3(x1), . . .
Ψ(x1, x2) =∞∑s=1
ϕs(x2)vs(x1)
if B is measured and found to be br , then the system can be said to be inthe state Ψ = ϕr (x2)vr (x1) and the second system must be in state ϕr (x2)
Thus, as a consequence of two different measurements made on System I,System II can be left in states with two different wave functions, when it isfar away from, and not interacting with System I
Thus one can assign two different wave functions, ψk(x2) and ϕr (x2), tothe same reality (System II after interaction with System I)
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 8 / 22
Einstein Podolsky Rosen paradox
If instead of operator A, we chooseto expand the overall wavefunctionin terms of the eigenfunctions of B,then we have
b1, b2, b3, . . .
v1(x1), v2(x1), v3(x1), . . .
Ψ(x1, x2) =∞∑s=1
ϕs(x2)vs(x1)
if B is measured and found to be br , then the system can be said to be inthe state Ψ = ϕr (x2)vr (x1) and the second system must be in state ϕr (x2)
Thus, as a consequence of two different measurements made on System I,System II can be left in states with two different wave functions, when it isfar away from, and not interacting with System I
Thus one can assign two different wave functions, ψk(x2) and ϕr (x2), tothe same reality (System II after interaction with System I)
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 8 / 22
Einstein Podolsky Rosen paradox
If instead of operator A, we chooseto expand the overall wavefunctionin terms of the eigenfunctions of B,then we have
b1, b2, b3, . . .
v1(x1), v2(x1), v3(x1), . . .
Ψ(x1, x2) =∞∑s=1
ϕs(x2)vs(x1)
if B is measured and found to be br , then the system can be said to be inthe state Ψ = ϕr (x2)vr (x1) and the second system must be in state ϕr (x2)
Thus, as a consequence of two different measurements made on System I,System II can be left in states with two different wave functions, when it isfar away from, and not interacting with System I
Thus one can assign two different wave functions, ψk(x2) and ϕr (x2), tothe same reality (System II after interaction with System I)
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 8 / 22
Einstein Podolsky Rosen paradox
Suppose that ψk(x2) and ϕr (x2) are eigenfunctions of two non-commutingoperators, P and Q with eigenvalues pk and qr
By measuring either A or B on System I, we are able to predict withcertainty, and without disturbing System II, either the value of P (pk) orQ (qr ) According to our criterion for reality, in the first case, P must bean element of reality and in the second case, Q must be an element ofreality but ψk(x2) and ϕr (x2) were shown to be part of the same realityand this leads to a contradiction of the postulate that “when the operatorscorresponding to two physical quantities do not commute the twoquantities cannot have simultaneous reality”
Thus, they conclude that “the quantum-mechanical description of realitygiven by the wave function is not complete”
If this “realist” version of quantum mechanics is correct, there must besome local hidden variables which specify the state of the systemcompletely
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 9 / 22
Einstein Podolsky Rosen paradox
Suppose that ψk(x2) and ϕr (x2) are eigenfunctions of two non-commutingoperators, P and Q with eigenvalues pk and qr
By measuring either A or B on System I, we are able to predict withcertainty, and without disturbing System II, either the value of P (pk) orQ (qr )
According to our criterion for reality, in the first case, P must bean element of reality and in the second case, Q must be an element ofreality but ψk(x2) and ϕr (x2) were shown to be part of the same realityand this leads to a contradiction of the postulate that “when the operatorscorresponding to two physical quantities do not commute the twoquantities cannot have simultaneous reality”
Thus, they conclude that “the quantum-mechanical description of realitygiven by the wave function is not complete”
If this “realist” version of quantum mechanics is correct, there must besome local hidden variables which specify the state of the systemcompletely
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 9 / 22
Einstein Podolsky Rosen paradox
Suppose that ψk(x2) and ϕr (x2) are eigenfunctions of two non-commutingoperators, P and Q with eigenvalues pk and qr
By measuring either A or B on System I, we are able to predict withcertainty, and without disturbing System II, either the value of P (pk) orQ (qr ) According to our criterion for reality, in the first case, P must bean element of reality and in the second case, Q must be an element ofreality but ψk(x2) and ϕr (x2) were shown to be part of the same realityand this leads to a contradiction of the postulate that “when the operatorscorresponding to two physical quantities do not commute the twoquantities cannot have simultaneous reality”
Thus, they conclude that “the quantum-mechanical description of realitygiven by the wave function is not complete”
If this “realist” version of quantum mechanics is correct, there must besome local hidden variables which specify the state of the systemcompletely
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 9 / 22
Einstein Podolsky Rosen paradox
Suppose that ψk(x2) and ϕr (x2) are eigenfunctions of two non-commutingoperators, P and Q with eigenvalues pk and qr
By measuring either A or B on System I, we are able to predict withcertainty, and without disturbing System II, either the value of P (pk) orQ (qr ) According to our criterion for reality, in the first case, P must bean element of reality and in the second case, Q must be an element ofreality but ψk(x2) and ϕr (x2) were shown to be part of the same realityand this leads to a contradiction of the postulate that “when the operatorscorresponding to two physical quantities do not commute the twoquantities cannot have simultaneous reality”
Thus, they conclude that “the quantum-mechanical description of realitygiven by the wave function is not complete”
If this “realist” version of quantum mechanics is correct, there must besome local hidden variables which specify the state of the systemcompletely
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 9 / 22
Einstein Podolsky Rosen paradox
Suppose that ψk(x2) and ϕr (x2) are eigenfunctions of two non-commutingoperators, P and Q with eigenvalues pk and qr
By measuring either A or B on System I, we are able to predict withcertainty, and without disturbing System II, either the value of P (pk) orQ (qr ) According to our criterion for reality, in the first case, P must bean element of reality and in the second case, Q must be an element ofreality but ψk(x2) and ϕr (x2) were shown to be part of the same realityand this leads to a contradiction of the postulate that “when the operatorscorresponding to two physical quantities do not commute the twoquantities cannot have simultaneous reality”
Thus, they conclude that “the quantum-mechanical description of realitygiven by the wave function is not complete”
If this “realist” version of quantum mechanics is correct, there must besome local hidden variables which specify the state of the systemcompletely
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 9 / 22
Problem 12.1
Consider a two-level system, |φa〉 and |φb〉, with 〈φi |φj〉 = δij
The two-particle state
α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉
cannot be expressed as a product |ψr (1)〉 |ψs(2)〉 for any one-particlestates |ψr 〉 and |ψs〉
Start by assuming
α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉 = |ψr (1)〉 |ψs(2)〉
expand |ψr 〉 and |ψs〉 in terms ofthe two basis states
|ψr 〉 = A |φa〉+ B |φb〉|ψs〉 = C |φa〉+ D |φb〉
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 10 / 22
Problem 12.1
Consider a two-level system, |φa〉 and |φb〉, with 〈φi |φj〉 = δij
The two-particle state
α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉
cannot be expressed as a product |ψr (1)〉 |ψs(2)〉 for any one-particlestates |ψr 〉 and |ψs〉
Start by assuming
α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉 = |ψr (1)〉 |ψs(2)〉
expand |ψr 〉 and |ψs〉 in terms ofthe two basis states
|ψr 〉 = A |φa〉+ B |φb〉|ψs〉 = C |φa〉+ D |φb〉
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 10 / 22
Problem 12.1
Consider a two-level system, |φa〉 and |φb〉, with 〈φi |φj〉 = δij
The two-particle state
α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉
cannot be expressed as a product |ψr (1)〉 |ψs(2)〉 for any one-particlestates |ψr 〉 and |ψs〉
Start by assuming
α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉 = |ψr (1)〉 |ψs(2)〉
expand |ψr 〉 and |ψs〉 in terms ofthe two basis states
|ψr 〉 = A |φa〉+ B |φb〉|ψs〉 = C |φa〉+ D |φb〉
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 10 / 22
Problem 12.1
Consider a two-level system, |φa〉 and |φb〉, with 〈φi |φj〉 = δij
The two-particle state
α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉
cannot be expressed as a product |ψr (1)〉 |ψs(2)〉 for any one-particlestates |ψr 〉 and |ψs〉
Start by assuming
α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉 = |ψr (1)〉 |ψs(2)〉
expand |ψr 〉 and |ψs〉 in terms ofthe two basis states
|ψr 〉 = A |φa〉+ B |φb〉|ψs〉 = C |φa〉+ D |φb〉
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 10 / 22
Problem 12.1
Consider a two-level system, |φa〉 and |φb〉, with 〈φi |φj〉 = δij
The two-particle state
α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉
cannot be expressed as a product |ψr (1)〉 |ψs(2)〉 for any one-particlestates |ψr 〉 and |ψs〉
Start by assuming
α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉 = |ψr (1)〉 |ψs(2)〉
expand |ψr 〉 and |ψs〉 in terms ofthe two basis states
|ψr 〉 = A |φa〉+ B |φb〉
|ψs〉 = C |φa〉+ D |φb〉
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 10 / 22
Problem 12.1
Consider a two-level system, |φa〉 and |φb〉, with 〈φi |φj〉 = δij
The two-particle state
α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉
cannot be expressed as a product |ψr (1)〉 |ψs(2)〉 for any one-particlestates |ψr 〉 and |ψs〉
Start by assuming
α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉 = |ψr (1)〉 |ψs(2)〉
expand |ψr 〉 and |ψs〉 in terms ofthe two basis states
|ψr 〉 = A |φa〉+ B |φb〉|ψs〉 = C |φa〉+ D |φb〉
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 10 / 22
Problem 12.1
α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉= [A |φa(1)〉+ B |φb(1)〉] [C |φa(2)〉+ D |φb(2)〉]
= AC |φa(1)〉 |φa(2)〉+ AD |φa(1)〉 |φb(2)〉+ BC |φb(1)〉 |φa(2)〉+ BD |φb(1)〉 |φb(2)〉
taking the inner products:
〈φa(1)| 〈φb(2)| −→ α = AD
〈φa(1)| 〈φa(2)| −→ 0 = AC
〈φb(1)| 〈φa(2)| −→ β = BC
〈φb(1)| 〈φb(2)| −→ 0 = BD
if any of the coefficients (A, B, C , or D) are zero, then either α or β mustbe zero which is a condition excluded by assumption
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 11 / 22
Problem 12.1
α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉= [A |φa(1)〉+ B |φb(1)〉] [C |φa(2)〉+ D |φb(2)〉]= AC |φa(1)〉 |φa(2)〉+ AD |φa(1)〉 |φb(2)〉
+ BC |φb(1)〉 |φa(2)〉+ BD |φb(1)〉 |φb(2)〉
taking the inner products:
〈φa(1)| 〈φb(2)| −→ α = AD
〈φa(1)| 〈φa(2)| −→ 0 = AC
〈φb(1)| 〈φa(2)| −→ β = BC
〈φb(1)| 〈φb(2)| −→ 0 = BD
if any of the coefficients (A, B, C , or D) are zero, then either α or β mustbe zero which is a condition excluded by assumption
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 11 / 22
Problem 12.1
α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉= [A |φa(1)〉+ B |φb(1)〉] [C |φa(2)〉+ D |φb(2)〉]= AC |φa(1)〉 |φa(2)〉+ AD |φa(1)〉 |φb(2)〉
+ BC |φb(1)〉 |φa(2)〉+ BD |φb(1)〉 |φb(2)〉
taking the inner products:
〈φa(1)| 〈φb(2)| −→ α = AD
〈φa(1)| 〈φa(2)| −→ 0 = AC
〈φb(1)| 〈φa(2)| −→ β = BC
〈φb(1)| 〈φb(2)| −→ 0 = BD
if any of the coefficients (A, B, C , or D) are zero, then either α or β mustbe zero which is a condition excluded by assumption
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 11 / 22
Problem 12.1
α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉= [A |φa(1)〉+ B |φb(1)〉] [C |φa(2)〉+ D |φb(2)〉]= AC |φa(1)〉 |φa(2)〉+ AD |φa(1)〉 |φb(2)〉
+ BC |φb(1)〉 |φa(2)〉+ BD |φb(1)〉 |φb(2)〉
taking the inner products:
〈φa(1)| 〈φb(2)|
−→ α = AD
〈φa(1)| 〈φa(2)| −→ 0 = AC
〈φb(1)| 〈φa(2)| −→ β = BC
〈φb(1)| 〈φb(2)| −→ 0 = BD
if any of the coefficients (A, B, C , or D) are zero, then either α or β mustbe zero which is a condition excluded by assumption
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 11 / 22
Problem 12.1
α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉= [A |φa(1)〉+ B |φb(1)〉] [C |φa(2)〉+ D |φb(2)〉]= AC |φa(1)〉 |φa(2)〉+ AD |φa(1)〉 |φb(2)〉
+ BC |φb(1)〉 |φa(2)〉+ BD |φb(1)〉 |φb(2)〉
taking the inner products:
〈φa(1)| 〈φb(2)| −→ α = AD
〈φa(1)| 〈φa(2)| −→ 0 = AC
〈φb(1)| 〈φa(2)| −→ β = BC
〈φb(1)| 〈φb(2)| −→ 0 = BD
if any of the coefficients (A, B, C , or D) are zero, then either α or β mustbe zero which is a condition excluded by assumption
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 11 / 22
Problem 12.1
α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉= [A |φa(1)〉+ B |φb(1)〉] [C |φa(2)〉+ D |φb(2)〉]= AC |φa(1)〉 |φa(2)〉+ AD |φa(1)〉 |φb(2)〉
+ BC |φb(1)〉 |φa(2)〉+ BD |φb(1)〉 |φb(2)〉
taking the inner products:
〈φa(1)| 〈φb(2)| −→ α = AD
〈φa(1)| 〈φa(2)|
−→ 0 = AC
〈φb(1)| 〈φa(2)| −→ β = BC
〈φb(1)| 〈φb(2)| −→ 0 = BD
if any of the coefficients (A, B, C , or D) are zero, then either α or β mustbe zero which is a condition excluded by assumption
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 11 / 22
Problem 12.1
α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉= [A |φa(1)〉+ B |φb(1)〉] [C |φa(2)〉+ D |φb(2)〉]= AC |φa(1)〉 |φa(2)〉+ AD |φa(1)〉 |φb(2)〉
+ BC |φb(1)〉 |φa(2)〉+ BD |φb(1)〉 |φb(2)〉
taking the inner products:
〈φa(1)| 〈φb(2)| −→ α = AD
〈φa(1)| 〈φa(2)| −→ 0 = AC
〈φb(1)| 〈φa(2)| −→ β = BC
〈φb(1)| 〈φb(2)| −→ 0 = BD
if any of the coefficients (A, B, C , or D) are zero, then either α or β mustbe zero which is a condition excluded by assumption
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 11 / 22
Problem 12.1
α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉= [A |φa(1)〉+ B |φb(1)〉] [C |φa(2)〉+ D |φb(2)〉]= AC |φa(1)〉 |φa(2)〉+ AD |φa(1)〉 |φb(2)〉
+ BC |φb(1)〉 |φa(2)〉+ BD |φb(1)〉 |φb(2)〉
taking the inner products:
〈φa(1)| 〈φb(2)| −→ α = AD
〈φa(1)| 〈φa(2)| −→ 0 = AC
〈φb(1)| 〈φa(2)|
−→ β = BC
〈φb(1)| 〈φb(2)| −→ 0 = BD
if any of the coefficients (A, B, C , or D) are zero, then either α or β mustbe zero which is a condition excluded by assumption
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 11 / 22
Problem 12.1
α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉= [A |φa(1)〉+ B |φb(1)〉] [C |φa(2)〉+ D |φb(2)〉]= AC |φa(1)〉 |φa(2)〉+ AD |φa(1)〉 |φb(2)〉
+ BC |φb(1)〉 |φa(2)〉+ BD |φb(1)〉 |φb(2)〉
taking the inner products:
〈φa(1)| 〈φb(2)| −→ α = AD
〈φa(1)| 〈φa(2)| −→ 0 = AC
〈φb(1)| 〈φa(2)| −→ β = BC
〈φb(1)| 〈φb(2)| −→ 0 = BD
if any of the coefficients (A, B, C , or D) are zero, then either α or β mustbe zero which is a condition excluded by assumption
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 11 / 22
Problem 12.1
α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉= [A |φa(1)〉+ B |φb(1)〉] [C |φa(2)〉+ D |φb(2)〉]= AC |φa(1)〉 |φa(2)〉+ AD |φa(1)〉 |φb(2)〉
+ BC |φb(1)〉 |φa(2)〉+ BD |φb(1)〉 |φb(2)〉
taking the inner products:
〈φa(1)| 〈φb(2)| −→ α = AD
〈φa(1)| 〈φa(2)| −→ 0 = AC
〈φb(1)| 〈φa(2)| −→ β = BC
〈φb(1)| 〈φb(2)|
−→ 0 = BD
if any of the coefficients (A, B, C , or D) are zero, then either α or β mustbe zero which is a condition excluded by assumption
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 11 / 22
Problem 12.1
α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉= [A |φa(1)〉+ B |φb(1)〉] [C |φa(2)〉+ D |φb(2)〉]= AC |φa(1)〉 |φa(2)〉+ AD |φa(1)〉 |φb(2)〉
+ BC |φb(1)〉 |φa(2)〉+ BD |φb(1)〉 |φb(2)〉
taking the inner products:
〈φa(1)| 〈φb(2)| −→ α = AD
〈φa(1)| 〈φa(2)| −→ 0 = AC
〈φb(1)| 〈φa(2)| −→ β = BC
〈φb(1)| 〈φb(2)| −→ 0 = BD
if any of the coefficients (A, B, C , or D) are zero, then either α or β mustbe zero which is a condition excluded by assumption
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 11 / 22
Problem 12.1
α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉= [A |φa(1)〉+ B |φb(1)〉] [C |φa(2)〉+ D |φb(2)〉]= AC |φa(1)〉 |φa(2)〉+ AD |φa(1)〉 |φb(2)〉
+ BC |φb(1)〉 |φa(2)〉+ BD |φb(1)〉 |φb(2)〉
taking the inner products:
〈φa(1)| 〈φb(2)| −→ α = AD
〈φa(1)| 〈φa(2)| −→ 0 = AC
〈φb(1)| 〈φa(2)| −→ β = BC
〈φb(1)| 〈φb(2)| −→ 0 = BD
if any of the coefficients (A, B, C , or D) are zero, then either α or β mustbe zero which is a condition excluded by assumption
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 11 / 22
Bell’s inequality
“On the Einstein Podolsky Rosen paradox,” J.S. Bell, Physics 1, 195-200(1964).
Consider an experiment wherein a pair of spin one-half particles areprepared in a singlet spin state and are moving freely in oppositedirections.
Suppose that two apparati are prepared to measure the spin state of eachparticle, ~σ1 and ~σ2 by means of Stern-Gerlach magnets which do notinterfere with the other particle.
If we measure the component ~σ1 · a and get +1 then we can presume that~σ2 · a = −1 without doing the measurement, the result is predetermined.
Assume that this predetermination is characterized by parameters λ.
The result, A, of measuring ~σ1 · a is then determined by a and λ and theresults, B, of measuring ~σ2 · b is then determined by b and λ.
A(a, λ) = ±1, B(b, λ) = ±1
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 12 / 22
Bell’s inequality
“On the Einstein Podolsky Rosen paradox,” J.S. Bell, Physics 1, 195-200(1964).
Consider an experiment wherein a pair of spin one-half particles areprepared in a singlet spin state and are moving freely in oppositedirections.
Suppose that two apparati are prepared to measure the spin state of eachparticle, ~σ1 and ~σ2 by means of Stern-Gerlach magnets which do notinterfere with the other particle.
If we measure the component ~σ1 · a and get +1 then we can presume that~σ2 · a = −1 without doing the measurement, the result is predetermined.
Assume that this predetermination is characterized by parameters λ.
The result, A, of measuring ~σ1 · a is then determined by a and λ and theresults, B, of measuring ~σ2 · b is then determined by b and λ.
A(a, λ) = ±1, B(b, λ) = ±1
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 12 / 22
Bell’s inequality
“On the Einstein Podolsky Rosen paradox,” J.S. Bell, Physics 1, 195-200(1964).
Consider an experiment wherein a pair of spin one-half particles areprepared in a singlet spin state and are moving freely in oppositedirections.
Suppose that two apparati are prepared to measure the spin state of eachparticle, ~σ1 and ~σ2 by means of Stern-Gerlach magnets which do notinterfere with the other particle.
If we measure the component ~σ1 · a and get +1 then we can presume that~σ2 · a = −1 without doing the measurement, the result is predetermined.
Assume that this predetermination is characterized by parameters λ.
The result, A, of measuring ~σ1 · a is then determined by a and λ and theresults, B, of measuring ~σ2 · b is then determined by b and λ.
A(a, λ) = ±1, B(b, λ) = ±1
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 12 / 22
Bell’s inequality
“On the Einstein Podolsky Rosen paradox,” J.S. Bell, Physics 1, 195-200(1964).
Consider an experiment wherein a pair of spin one-half particles areprepared in a singlet spin state and are moving freely in oppositedirections.
Suppose that two apparati are prepared to measure the spin state of eachparticle, ~σ1 and ~σ2 by means of Stern-Gerlach magnets which do notinterfere with the other particle.
If we measure the component ~σ1 · a and get +1
then we can presume that~σ2 · a = −1 without doing the measurement, the result is predetermined.
Assume that this predetermination is characterized by parameters λ.
The result, A, of measuring ~σ1 · a is then determined by a and λ and theresults, B, of measuring ~σ2 · b is then determined by b and λ.
A(a, λ) = ±1, B(b, λ) = ±1
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 12 / 22
Bell’s inequality
“On the Einstein Podolsky Rosen paradox,” J.S. Bell, Physics 1, 195-200(1964).
Consider an experiment wherein a pair of spin one-half particles areprepared in a singlet spin state and are moving freely in oppositedirections.
Suppose that two apparati are prepared to measure the spin state of eachparticle, ~σ1 and ~σ2 by means of Stern-Gerlach magnets which do notinterfere with the other particle.
If we measure the component ~σ1 · a and get +1 then we can presume that~σ2 · a = −1 without doing the measurement, the result is predetermined.
Assume that this predetermination is characterized by parameters λ.
The result, A, of measuring ~σ1 · a is then determined by a and λ and theresults, B, of measuring ~σ2 · b is then determined by b and λ.
A(a, λ) = ±1, B(b, λ) = ±1
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 12 / 22
Bell’s inequality
“On the Einstein Podolsky Rosen paradox,” J.S. Bell, Physics 1, 195-200(1964).
Consider an experiment wherein a pair of spin one-half particles areprepared in a singlet spin state and are moving freely in oppositedirections.
Suppose that two apparati are prepared to measure the spin state of eachparticle, ~σ1 and ~σ2 by means of Stern-Gerlach magnets which do notinterfere with the other particle.
If we measure the component ~σ1 · a and get +1 then we can presume that~σ2 · a = −1 without doing the measurement, the result is predetermined.
Assume that this predetermination is characterized by parameters λ.
The result, A, of measuring ~σ1 · a is then determined by a and λ and theresults, B, of measuring ~σ2 · b is then determined by b and λ.
A(a, λ) = ±1, B(b, λ) = ±1
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 12 / 22
Bell’s inequality
“On the Einstein Podolsky Rosen paradox,” J.S. Bell, Physics 1, 195-200(1964).
Consider an experiment wherein a pair of spin one-half particles areprepared in a singlet spin state and are moving freely in oppositedirections.
Suppose that two apparati are prepared to measure the spin state of eachparticle, ~σ1 and ~σ2 by means of Stern-Gerlach magnets which do notinterfere with the other particle.
If we measure the component ~σ1 · a and get +1 then we can presume that~σ2 · a = −1 without doing the measurement, the result is predetermined.
Assume that this predetermination is characterized by parameters λ.
The result, A, of measuring ~σ1 · a is then determined by a and λ
and theresults, B, of measuring ~σ2 · b is then determined by b and λ.
A(a, λ) = ±1,
B(b, λ) = ±1
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 12 / 22
Bell’s inequality
“On the Einstein Podolsky Rosen paradox,” J.S. Bell, Physics 1, 195-200(1964).
Consider an experiment wherein a pair of spin one-half particles areprepared in a singlet spin state and are moving freely in oppositedirections.
Suppose that two apparati are prepared to measure the spin state of eachparticle, ~σ1 and ~σ2 by means of Stern-Gerlach magnets which do notinterfere with the other particle.
If we measure the component ~σ1 · a and get +1 then we can presume that~σ2 · a = −1 without doing the measurement, the result is predetermined.
Assume that this predetermination is characterized by parameters λ.
The result, A, of measuring ~σ1 · a is then determined by a and λ and theresults, B, of measuring ~σ2 · b is then determined by b and λ.
A(a, λ) = ±1, B(b, λ) = ±1
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 12 / 22
Bell’s inequality
Result B for particle 2 is assumed not to depend on a and neither does Adepend on b.
If ρ(λ) is the probability distribu-tion of λ, then the expectationvalue of the product of ~σ1 · a and~σ2 · b is
and it should be equivalent to thequantum expectation value
P(a, b) =
∫ρ(λ)A(a, λ)B(b, λ) dλ
〈~σ1 · a |~σ2 · b〉 = −a · b
The essence of Bell’s paper is that this last statement is not possiblebecause of the presence of the hidden local variable(s) λ
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 13 / 22
Bell’s inequality
Result B for particle 2 is assumed not to depend on a and neither does Adepend on b.
If ρ(λ) is the probability distribu-tion of λ, then the expectationvalue of the product of ~σ1 · a and~σ2 · b is
and it should be equivalent to thequantum expectation value
P(a, b) =
∫ρ(λ)A(a, λ)B(b, λ) dλ
〈~σ1 · a |~σ2 · b〉 = −a · b
The essence of Bell’s paper is that this last statement is not possiblebecause of the presence of the hidden local variable(s) λ
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 13 / 22
Bell’s inequality
Result B for particle 2 is assumed not to depend on a and neither does Adepend on b.
If ρ(λ) is the probability distribu-tion of λ, then the expectationvalue of the product of ~σ1 · a and~σ2 · b is
and it should be equivalent to thequantum expectation value
P(a, b) =
∫ρ(λ)A(a, λ)B(b, λ) dλ
〈~σ1 · a |~σ2 · b〉 = −a · b
The essence of Bell’s paper is that this last statement is not possiblebecause of the presence of the hidden local variable(s) λ
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 13 / 22
Bell’s inequality
Result B for particle 2 is assumed not to depend on a and neither does Adepend on b.
If ρ(λ) is the probability distribu-tion of λ, then the expectationvalue of the product of ~σ1 · a and~σ2 · b is
and it should be equivalent to thequantum expectation value
P(a, b) =
∫ρ(λ)A(a, λ)B(b, λ) dλ
〈~σ1 · a |~σ2 · b〉 = −a · b
The essence of Bell’s paper is that this last statement is not possiblebecause of the presence of the hidden local variable(s) λ
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 13 / 22
Bell’s inequality
Result B for particle 2 is assumed not to depend on a and neither does Adepend on b.
If ρ(λ) is the probability distribu-tion of λ, then the expectationvalue of the product of ~σ1 · a and~σ2 · b is
and it should be equivalent to thequantum expectation value
P(a, b) =
∫ρ(λ)A(a, λ)B(b, λ) dλ
〈~σ1 · a |~σ2 · b〉 = −a · b
The essence of Bell’s paper is that this last statement is not possiblebecause of the presence of the hidden local variable(s) λ
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 13 / 22
Bell’s inequality
Result B for particle 2 is assumed not to depend on a and neither does Adepend on b.
If ρ(λ) is the probability distribu-tion of λ, then the expectationvalue of the product of ~σ1 · a and~σ2 · b is
and it should be equivalent to thequantum expectation value
P(a, b) =
∫ρ(λ)A(a, λ)B(b, λ) dλ
〈~σ1 · a |~σ2 · b〉 = −a · b
The essence of Bell’s paper is that this last statement is not possiblebecause of the presence of the hidden local variable(s) λ
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 13 / 22
Aside: Problem 4.50
Suppose we have 2 spin 12 fermions in a singlet configuration. If Sa1 is the
component of the first particle’s spin angular momentum in the direction aand Sb2 is the component of the second particle’s spin angular momentumin the b direction. Compute 〈Sa1Sb2〉
The singlet state can be written as
if we choose a ≡ z and put b in thex-z plane
|0 0〉 =1√2
(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
Sa1 = Sz1
Sb2 = cos θSz2 + sin θSx2
Sa1Sb2 |0 0〉 =
1√2
[Sz1(cos θSz2 + sin θSx2)] (|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
=1√2
[Sz1 |↑1〉 (cos θSz2 |↓2〉
+ sin θSx2 |↓2〉
)
− Sz1 |↓1〉 (cos θSz2 |↑2〉
+ sin θSx2 |↑2〉
)
]
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 14 / 22
Aside: Problem 4.50
Suppose we have 2 spin 12 fermions in a singlet configuration. If Sa1 is the
component of the first particle’s spin angular momentum in the direction aand Sb2 is the component of the second particle’s spin angular momentumin the b direction. Compute 〈Sa1Sb2〉
The singlet state can be written as
if we choose a ≡ z and put b in thex-z plane
|0 0〉 =1√2
(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
Sa1 = Sz1
Sb2 = cos θSz2 + sin θSx2
Sa1Sb2 |0 0〉 =
1√2
[Sz1(cos θSz2 + sin θSx2)] (|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
=1√2
[Sz1 |↑1〉 (cos θSz2 |↓2〉
+ sin θSx2 |↓2〉
)
− Sz1 |↓1〉 (cos θSz2 |↑2〉
+ sin θSx2 |↑2〉
)
]
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 14 / 22
Aside: Problem 4.50
Suppose we have 2 spin 12 fermions in a singlet configuration. If Sa1 is the
component of the first particle’s spin angular momentum in the direction aand Sb2 is the component of the second particle’s spin angular momentumin the b direction. Compute 〈Sa1Sb2〉
The singlet state can be written as
if we choose a ≡ z and put b in thex-z plane
|0 0〉 =1√2
(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
Sa1 = Sz1
Sb2 = cos θSz2 + sin θSx2
Sa1Sb2 |0 0〉 =
1√2
[Sz1(cos θSz2 + sin θSx2)] (|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
=1√2
[Sz1 |↑1〉 (cos θSz2 |↓2〉
+ sin θSx2 |↓2〉
)
− Sz1 |↓1〉 (cos θSz2 |↑2〉
+ sin θSx2 |↑2〉
)
]
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 14 / 22
Aside: Problem 4.50
Suppose we have 2 spin 12 fermions in a singlet configuration. If Sa1 is the
component of the first particle’s spin angular momentum in the direction aand Sb2 is the component of the second particle’s spin angular momentumin the b direction. Compute 〈Sa1Sb2〉
The singlet state can be written as
if we choose a ≡ z and put b in thex-z plane
|0 0〉 =1√2
(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
Sa1 = Sz1
Sb2 = cos θSz2 + sin θSx2
Sa1Sb2 |0 0〉 =
1√2
[Sz1(cos θSz2 + sin θSx2)] (|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
=1√2
[Sz1 |↑1〉 (cos θSz2 |↓2〉
+ sin θSx2 |↓2〉
)
− Sz1 |↓1〉 (cos θSz2 |↑2〉
+ sin θSx2 |↑2〉
)
]
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 14 / 22
Aside: Problem 4.50
Suppose we have 2 spin 12 fermions in a singlet configuration. If Sa1 is the
component of the first particle’s spin angular momentum in the direction aand Sb2 is the component of the second particle’s spin angular momentumin the b direction. Compute 〈Sa1Sb2〉
The singlet state can be written as
if we choose a ≡ z and put b in thex-z plane
|0 0〉 =1√2
(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
Sa1 = Sz1
Sb2 = cos θSz2 + sin θSx2
Sa1Sb2 |0 0〉 =
1√2
[Sz1(cos θSz2 + sin θSx2)] (|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
=1√2
[Sz1 |↑1〉 (cos θSz2 |↓2〉
+ sin θSx2 |↓2〉
)
− Sz1 |↓1〉 (cos θSz2 |↑2〉
+ sin θSx2 |↑2〉
)
]
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 14 / 22
Aside: Problem 4.50
Suppose we have 2 spin 12 fermions in a singlet configuration. If Sa1 is the
component of the first particle’s spin angular momentum in the direction aand Sb2 is the component of the second particle’s spin angular momentumin the b direction. Compute 〈Sa1Sb2〉
The singlet state can be written as
if we choose a ≡ z and put b in thex-z plane
|0 0〉 =1√2
(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
Sa1 = Sz1
Sb2 = cos θSz2 + sin θSx2
Sa1Sb2 |0 0〉 =
1√2
[Sz1(cos θSz2 + sin θSx2)] (|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
=1√2
[Sz1 |↑1〉 (cos θSz2 |↓2〉
+ sin θSx2 |↓2〉
)
− Sz1 |↓1〉 (cos θSz2 |↑2〉
+ sin θSx2 |↑2〉
)
]
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 14 / 22
Aside: Problem 4.50
Suppose we have 2 spin 12 fermions in a singlet configuration. If Sa1 is the
component of the first particle’s spin angular momentum in the direction aand Sb2 is the component of the second particle’s spin angular momentumin the b direction. Compute 〈Sa1Sb2〉
The singlet state can be written as
if we choose a ≡ z and put b in thex-z plane
|0 0〉 =1√2
(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
Sa1 = Sz1
Sb2 = cos θSz2 + sin θSx2
Sa1Sb2 |0 0〉 =
1√2
[Sz1(cos θSz2 + sin θSx2)] (|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
=1√2
[Sz1 |↑1〉 (cos θSz2 |↓2〉
+ sin θSx2 |↓2〉
)
− Sz1 |↓1〉 (cos θSz2 |↑2〉
+ sin θSx2 |↑2〉
)
]
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 14 / 22
Aside: Problem 4.50
Suppose we have 2 spin 12 fermions in a singlet configuration. If Sa1 is the
component of the first particle’s spin angular momentum in the direction aand Sb2 is the component of the second particle’s spin angular momentumin the b direction. Compute 〈Sa1Sb2〉
The singlet state can be written as
if we choose a ≡ z and put b in thex-z plane
|0 0〉 =1√2
(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
Sa1 = Sz1
Sb2 = cos θSz2 + sin θSx2
Sa1Sb2 |0 0〉 =1√2
[Sz1(cos θSz2 + sin θSx2)] (|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
=1√2
[Sz1 |↑1〉 (cos θSz2 |↓2〉
+ sin θSx2 |↓2〉
)
− Sz1 |↓1〉 (cos θSz2 |↑2〉
+ sin θSx2 |↑2〉
)
]
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 14 / 22
Aside: Problem 4.50
Suppose we have 2 spin 12 fermions in a singlet configuration. If Sa1 is the
component of the first particle’s spin angular momentum in the direction aand Sb2 is the component of the second particle’s spin angular momentumin the b direction. Compute 〈Sa1Sb2〉
The singlet state can be written as
if we choose a ≡ z and put b in thex-z plane
|0 0〉 =1√2
(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
Sa1 = Sz1
Sb2 = cos θSz2 + sin θSx2
Sa1Sb2 |0 0〉 =1√2
[Sz1(cos θSz2 + sin θSx2)] (|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
=1√2
[Sz1 |↑1〉 (cos θSz2 |↓2〉
+ sin θSx2 |↓2〉
)
− Sz1 |↓1〉 (cos θSz2 |↑2〉
+ sin θSx2 |↑2〉
)
]
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 14 / 22
Aside: Problem 4.50
Suppose we have 2 spin 12 fermions in a singlet configuration. If Sa1 is the
component of the first particle’s spin angular momentum in the direction aand Sb2 is the component of the second particle’s spin angular momentumin the b direction. Compute 〈Sa1Sb2〉
The singlet state can be written as
if we choose a ≡ z and put b in thex-z plane
|0 0〉 =1√2
(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
Sa1 = Sz1
Sb2 = cos θSz2 + sin θSx2
Sa1Sb2 |0 0〉 =1√2
[Sz1(cos θSz2 + sin θSx2)] (|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
=1√2
[Sz1 |↑1〉 (cos θSz2 |↓2〉 + sin θSx2 |↓2〉 )
− Sz1 |↓1〉 (cos θSz2 |↑2〉
+ sin θSx2 |↑2〉
)
]
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 14 / 22
Aside: Problem 4.50
Suppose we have 2 spin 12 fermions in a singlet configuration. If Sa1 is the
component of the first particle’s spin angular momentum in the direction aand Sb2 is the component of the second particle’s spin angular momentumin the b direction. Compute 〈Sa1Sb2〉
The singlet state can be written as
if we choose a ≡ z and put b in thex-z plane
|0 0〉 =1√2
(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
Sa1 = Sz1
Sb2 = cos θSz2 + sin θSx2
Sa1Sb2 |0 0〉 =1√2
[Sz1(cos θSz2 + sin θSx2)] (|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
=1√2
[Sz1 |↑1〉 (cos θSz2 |↓2〉 + sin θSx2 |↓2〉 )
− Sz1 |↓1〉 (cos θSz2 |↑2〉
+ sin θSx2 |↑2〉
) ]
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 14 / 22
Aside: Problem 4.50
Suppose we have 2 spin 12 fermions in a singlet configuration. If Sa1 is the
component of the first particle’s spin angular momentum in the direction aand Sb2 is the component of the second particle’s spin angular momentumin the b direction. Compute 〈Sa1Sb2〉
The singlet state can be written as
if we choose a ≡ z and put b in thex-z plane
|0 0〉 =1√2
(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
Sa1 = Sz1
Sb2 = cos θSz2 + sin θSx2
Sa1Sb2 |0 0〉 =1√2
[Sz1(cos θSz2 + sin θSx2)] (|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
=1√2
[Sz1 |↑1〉 (cos θSz2 |↓2〉 + sin θSx2 |↓2〉 )
− Sz1 |↓1〉 (cos θSz2 |↑2〉 + sin θSx2 |↑2〉 ) ]
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 14 / 22
Aside: Problem 4.50
Sa1Sb2 |0 0〉 =1√2
[Sz1 |↑1〉 (cos θSz2 |↓2〉+ sin θSx2 |↓2〉)
− Sz1 |↓1〉 (cos θSz2 |↑2〉+ sin θSx2 |↑2〉) ]
=1√2
~2
4
{|↑1〉 [− cos θ |↓2〉+ sin θ |↑2〉]
+ |↓1〉 [cos θ |↑2〉+ sin θ |↓2〉]}
=~2
4
[− cos θ
1√2
(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
+ sin θ1√2
(|↑1〉 |↑2〉+ |↓1〉 |↓2〉)]
〈0 0|Sa1Sb2 |0 0〉 = −~2
4cos θ 〈0 0 | 0 0〉 = −~2
4cos θ = −~2
4a · b
putting this in terms of the Pauli spin matrices: 〈~σ1 · a |~σ2 · b〉 = −a · b
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 15 / 22
Aside: Problem 4.50
Sa1Sb2 |0 0〉 =1√2
[Sz1 |↑1〉 (cos θSz2 |↓2〉+ sin θSx2 |↓2〉)
− Sz1 |↓1〉 (cos θSz2 |↑2〉+ sin θSx2 |↑2〉) ]
=1√2
~2
4
{|↑1〉 [− cos θ |↓2〉+ sin θ |↑2〉]
+ |↓1〉 [cos θ |↑2〉+ sin θ |↓2〉]}
=~2
4
[− cos θ
1√2
(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
+ sin θ1√2
(|↑1〉 |↑2〉+ |↓1〉 |↓2〉)]
〈0 0|Sa1Sb2 |0 0〉 = −~2
4cos θ 〈0 0 | 0 0〉 = −~2
4cos θ = −~2
4a · b
putting this in terms of the Pauli spin matrices: 〈~σ1 · a |~σ2 · b〉 = −a · b
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 15 / 22
Aside: Problem 4.50
Sa1Sb2 |0 0〉 =1√2
[Sz1 |↑1〉 (cos θSz2 |↓2〉+ sin θSx2 |↓2〉)
− Sz1 |↓1〉 (cos θSz2 |↑2〉+ sin θSx2 |↑2〉) ]
=1√2
~2
4
{|↑1〉 [− cos θ |↓2〉+ sin θ |↑2〉]
+ |↓1〉 [cos θ |↑2〉+ sin θ |↓2〉]}
=~2
4
[− cos θ
1√2
(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
+ sin θ1√2
(|↑1〉 |↑2〉+ |↓1〉 |↓2〉)]
〈0 0|Sa1Sb2 |0 0〉 = −~2
4cos θ 〈0 0 | 0 0〉 = −~2
4cos θ = −~2
4a · b
putting this in terms of the Pauli spin matrices: 〈~σ1 · a |~σ2 · b〉 = −a · b
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 15 / 22
Aside: Problem 4.50
Sa1Sb2 |0 0〉 =1√2
[Sz1 |↑1〉 (cos θSz2 |↓2〉+ sin θSx2 |↓2〉)
− Sz1 |↓1〉 (cos θSz2 |↑2〉+ sin θSx2 |↑2〉) ]
=1√2
~2
4
{|↑1〉 [− cos θ |↓2〉+ sin θ |↑2〉]
+ |↓1〉 [cos θ |↑2〉+ sin θ |↓2〉]}
=~2
4
[− cos θ
1√2
(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
+ sin θ1√2
(|↑1〉 |↑2〉+ |↓1〉 |↓2〉)]
〈0 0| Sa1Sb2 |0 0〉 = −~2
4cos θ 〈0 0 | 0 0〉
= −~2
4cos θ = −~2
4a · b
putting this in terms of the Pauli spin matrices: 〈~σ1 · a |~σ2 · b〉 = −a · b
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 15 / 22
Aside: Problem 4.50
Sa1Sb2 |0 0〉 =1√2
[Sz1 |↑1〉 (cos θSz2 |↓2〉+ sin θSx2 |↓2〉)
− Sz1 |↓1〉 (cos θSz2 |↑2〉+ sin θSx2 |↑2〉) ]
=1√2
~2
4
{|↑1〉 [− cos θ |↓2〉+ sin θ |↑2〉]
+ |↓1〉 [cos θ |↑2〉+ sin θ |↓2〉]}
=~2
4
[− cos θ
1√2
(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
+ sin θ1√2
(|↑1〉 |↑2〉+ |↓1〉 |↓2〉)]
〈0 0| Sa1Sb2 |0 0〉 = −~2
4cos θ 〈0 0 | 0 0〉 = −~2
4cos θ
= −~2
4a · b
putting this in terms of the Pauli spin matrices: 〈~σ1 · a |~σ2 · b〉 = −a · b
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 15 / 22
Aside: Problem 4.50
Sa1Sb2 |0 0〉 =1√2
[Sz1 |↑1〉 (cos θSz2 |↓2〉+ sin θSx2 |↓2〉)
− Sz1 |↓1〉 (cos θSz2 |↑2〉+ sin θSx2 |↑2〉) ]
=1√2
~2
4
{|↑1〉 [− cos θ |↓2〉+ sin θ |↑2〉]
+ |↓1〉 [cos θ |↑2〉+ sin θ |↓2〉]}
=~2
4
[− cos θ
1√2
(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
+ sin θ1√2
(|↑1〉 |↑2〉+ |↓1〉 |↓2〉)]
〈0 0| Sa1Sb2 |0 0〉 = −~2
4cos θ 〈0 0 | 0 0〉 = −~2
4cos θ = −~2
4a · b
putting this in terms of the Pauli spin matrices: 〈~σ1 · a |~σ2 · b〉 = −a · b
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 15 / 22
Aside: Problem 4.50
Sa1Sb2 |0 0〉 =1√2
[Sz1 |↑1〉 (cos θSz2 |↓2〉+ sin θSx2 |↓2〉)
− Sz1 |↓1〉 (cos θSz2 |↑2〉+ sin θSx2 |↑2〉) ]
=1√2
~2
4
{|↑1〉 [− cos θ |↓2〉+ sin θ |↑2〉]
+ |↓1〉 [cos θ |↑2〉+ sin θ |↓2〉]}
=~2
4
[− cos θ
1√2
(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)
+ sin θ1√2
(|↑1〉 |↑2〉+ |↓1〉 |↓2〉)]
〈0 0| Sa1Sb2 |0 0〉 = −~2
4cos θ 〈0 0 | 0 0〉 = −~2
4cos θ = −~2
4a · b
putting this in terms of the Pauli spin matrices: 〈~σ1 · a |~σ2 · b〉 = −a · b
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 15 / 22
Bell’s inequality
Assuming the probability distribu-tion ρ(λ) is normalized
we note that becauseA(a, λ) = ±1, B(b, λ) = ±1
∫ρ(λ) dλ = 1
P(a, b) ≥ −1
P(a, b) =
∫ρ(λ)A(a, λ)B(b, λ) dλ = −
∫ρ(λ)A(a, λ)A(b, λ) dλ
P(a, b) can only reach -1 when
defining c as a third unit vectora = b
A(a, λ) = −B(a, λ)
P(a, b)− P(a, c) = −∫ρ(λ)
[A(a, λ)A(b, λ)− A(a, λ)A(c , λ)
]dλ
= −∫ρ(λ)
[A(a, λ)A(b, λ)− A(b, λ)A(b, λ)A(a, λ)A(c , λ)
]dλ
= −∫ρ(λ)A(a, λ)A(b, λ)
[1− A(b, λ)A(c , λ)
]dλ
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 16 / 22
Bell’s inequality
Assuming the probability distribu-tion ρ(λ) is normalized
we note that becauseA(a, λ) = ±1, B(b, λ) = ±1
∫ρ(λ) dλ = 1
P(a, b) ≥ −1
P(a, b) =
∫ρ(λ)A(a, λ)B(b, λ) dλ = −
∫ρ(λ)A(a, λ)A(b, λ) dλ
P(a, b) can only reach -1 when
defining c as a third unit vectora = b
A(a, λ) = −B(a, λ)
P(a, b)− P(a, c) = −∫ρ(λ)
[A(a, λ)A(b, λ)− A(a, λ)A(c , λ)
]dλ
= −∫ρ(λ)
[A(a, λ)A(b, λ)− A(b, λ)A(b, λ)A(a, λ)A(c , λ)
]dλ
= −∫ρ(λ)A(a, λ)A(b, λ)
[1− A(b, λ)A(c , λ)
]dλ
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 16 / 22
Bell’s inequality
Assuming the probability distribu-tion ρ(λ) is normalized
we note that becauseA(a, λ) = ±1, B(b, λ) = ±1
∫ρ(λ) dλ = 1
P(a, b) ≥ −1
P(a, b) =
∫ρ(λ)A(a, λ)B(b, λ) dλ = −
∫ρ(λ)A(a, λ)A(b, λ) dλ
P(a, b) can only reach -1 when
defining c as a third unit vectora = b
A(a, λ) = −B(a, λ)
P(a, b)− P(a, c) = −∫ρ(λ)
[A(a, λ)A(b, λ)− A(a, λ)A(c , λ)
]dλ
= −∫ρ(λ)
[A(a, λ)A(b, λ)− A(b, λ)A(b, λ)A(a, λ)A(c , λ)
]dλ
= −∫ρ(λ)A(a, λ)A(b, λ)
[1− A(b, λ)A(c , λ)
]dλ
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 16 / 22
Bell’s inequality
Assuming the probability distribu-tion ρ(λ) is normalized
we note that becauseA(a, λ) = ±1, B(b, λ) = ±1
∫ρ(λ) dλ = 1
P(a, b) ≥ −1
P(a, b) =
∫ρ(λ)A(a, λ)B(b, λ) dλ = −
∫ρ(λ)A(a, λ)A(b, λ) dλ
P(a, b) can only reach -1 when
defining c as a third unit vectora = b
A(a, λ) = −B(a, λ)
P(a, b)− P(a, c) = −∫ρ(λ)
[A(a, λ)A(b, λ)− A(a, λ)A(c , λ)
]dλ
= −∫ρ(λ)
[A(a, λ)A(b, λ)− A(b, λ)A(b, λ)A(a, λ)A(c , λ)
]dλ
= −∫ρ(λ)A(a, λ)A(b, λ)
[1− A(b, λ)A(c , λ)
]dλ
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 16 / 22
Bell’s inequality
Assuming the probability distribu-tion ρ(λ) is normalized
we note that becauseA(a, λ) = ±1, B(b, λ) = ±1
∫ρ(λ) dλ = 1
P(a, b) ≥ −1
P(a, b) =
∫ρ(λ)A(a, λ)B(b, λ) dλ
= −∫ρ(λ)A(a, λ)A(b, λ) dλ
P(a, b) can only reach -1 when
defining c as a third unit vectora = b
A(a, λ) = −B(a, λ)
P(a, b)− P(a, c) = −∫ρ(λ)
[A(a, λ)A(b, λ)− A(a, λ)A(c , λ)
]dλ
= −∫ρ(λ)
[A(a, λ)A(b, λ)− A(b, λ)A(b, λ)A(a, λ)A(c , λ)
]dλ
= −∫ρ(λ)A(a, λ)A(b, λ)
[1− A(b, λ)A(c , λ)
]dλ
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 16 / 22
Bell’s inequality
Assuming the probability distribu-tion ρ(λ) is normalized
we note that becauseA(a, λ) = ±1, B(b, λ) = ±1
∫ρ(λ) dλ = 1
P(a, b) ≥ −1
P(a, b) =
∫ρ(λ)A(a, λ)B(b, λ) dλ
= −∫ρ(λ)A(a, λ)A(b, λ) dλ
P(a, b) can only reach -1 when
defining c as a third unit vectora = b
A(a, λ) = −B(a, λ)
P(a, b)− P(a, c) = −∫ρ(λ)
[A(a, λ)A(b, λ)− A(a, λ)A(c , λ)
]dλ
= −∫ρ(λ)
[A(a, λ)A(b, λ)− A(b, λ)A(b, λ)A(a, λ)A(c , λ)
]dλ
= −∫ρ(λ)A(a, λ)A(b, λ)
[1− A(b, λ)A(c , λ)
]dλ
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 16 / 22
Bell’s inequality
Assuming the probability distribu-tion ρ(λ) is normalized
we note that becauseA(a, λ) = ±1, B(b, λ) = ±1
∫ρ(λ) dλ = 1
P(a, b) ≥ −1
P(a, b) =
∫ρ(λ)A(a, λ)B(b, λ) dλ
= −∫ρ(λ)A(a, λ)A(b, λ) dλ
P(a, b) can only reach -1 when
defining c as a third unit vector
a = b
A(a, λ) = −B(a, λ)
P(a, b)− P(a, c) = −∫ρ(λ)
[A(a, λ)A(b, λ)− A(a, λ)A(c , λ)
]dλ
= −∫ρ(λ)
[A(a, λ)A(b, λ)− A(b, λ)A(b, λ)A(a, λ)A(c , λ)
]dλ
= −∫ρ(λ)A(a, λ)A(b, λ)
[1− A(b, λ)A(c , λ)
]dλ
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 16 / 22
Bell’s inequality
Assuming the probability distribu-tion ρ(λ) is normalized
we note that becauseA(a, λ) = ±1, B(b, λ) = ±1
∫ρ(λ) dλ = 1
P(a, b) ≥ −1
P(a, b) =
∫ρ(λ)A(a, λ)B(b, λ) dλ
= −∫ρ(λ)A(a, λ)A(b, λ) dλ
P(a, b) can only reach -1 when
defining c as a third unit vector
a = b
A(a, λ) = −B(a, λ)
P(a, b)− P(a, c) = −∫ρ(λ)
[A(a, λ)A(b, λ)− A(a, λ)A(c , λ)
]dλ
= −∫ρ(λ)
[A(a, λ)A(b, λ)− A(b, λ)A(b, λ)A(a, λ)A(c , λ)
]dλ
= −∫ρ(λ)A(a, λ)A(b, λ)
[1− A(b, λ)A(c , λ)
]dλ
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 16 / 22
Bell’s inequality
Assuming the probability distribu-tion ρ(λ) is normalized
we note that becauseA(a, λ) = ±1, B(b, λ) = ±1
∫ρ(λ) dλ = 1
P(a, b) ≥ −1
P(a, b) =
∫ρ(λ)A(a, λ)B(b, λ) dλ = −
∫ρ(λ)A(a, λ)A(b, λ) dλ
P(a, b) can only reach -1 when
defining c as a third unit vector
a = b
A(a, λ) = −B(a, λ)
P(a, b)− P(a, c) = −∫ρ(λ)
[A(a, λ)A(b, λ)− A(a, λ)A(c , λ)
]dλ
= −∫ρ(λ)
[A(a, λ)A(b, λ)− A(b, λ)A(b, λ)A(a, λ)A(c , λ)
]dλ
= −∫ρ(λ)A(a, λ)A(b, λ)
[1− A(b, λ)A(c , λ)
]dλ
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 16 / 22
Bell’s inequality
Assuming the probability distribu-tion ρ(λ) is normalized
we note that becauseA(a, λ) = ±1, B(b, λ) = ±1
∫ρ(λ) dλ = 1
P(a, b) ≥ −1
P(a, b) =
∫ρ(λ)A(a, λ)B(b, λ) dλ = −
∫ρ(λ)A(a, λ)A(b, λ) dλ
P(a, b) can only reach -1 when
defining c as a third unit vectora = b
A(a, λ) = −B(a, λ)
P(a, b)− P(a, c) = −∫ρ(λ)
[A(a, λ)A(b, λ)− A(a, λ)A(c , λ)
]dλ
= −∫ρ(λ)
[A(a, λ)A(b, λ)− A(b, λ)A(b, λ)A(a, λ)A(c , λ)
]dλ
= −∫ρ(λ)A(a, λ)A(b, λ)
[1− A(b, λ)A(c , λ)
]dλ
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 16 / 22
Bell’s inequality
Assuming the probability distribu-tion ρ(λ) is normalized
we note that becauseA(a, λ) = ±1, B(b, λ) = ±1
∫ρ(λ) dλ = 1
P(a, b) ≥ −1
P(a, b) =
∫ρ(λ)A(a, λ)B(b, λ) dλ = −
∫ρ(λ)A(a, λ)A(b, λ) dλ
P(a, b) can only reach -1 when
defining c as a third unit vectora = b
A(a, λ) = −B(a, λ)
P(a, b)− P(a, c) = −∫ρ(λ)
[A(a, λ)A(b, λ)− A(a, λ)A(c , λ)
]dλ
= −∫ρ(λ)
[A(a, λ)A(b, λ)− A(b, λ)A(b, λ)A(a, λ)A(c , λ)
]dλ
= −∫ρ(λ)A(a, λ)A(b, λ)
[1− A(b, λ)A(c , λ)
]dλ
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 16 / 22
Bell’s inequality
Assuming the probability distribu-tion ρ(λ) is normalized
we note that becauseA(a, λ) = ±1, B(b, λ) = ±1
∫ρ(λ) dλ = 1
P(a, b) ≥ −1
P(a, b) =
∫ρ(λ)A(a, λ)B(b, λ) dλ = −
∫ρ(λ)A(a, λ)A(b, λ) dλ
P(a, b) can only reach -1 when
defining c as a third unit vectora = b
A(a, λ) = −B(a, λ)
P(a, b)− P(a, c) = −∫ρ(λ)
[A(a, λ)A(b, λ)− A(a, λ)A(c , λ)
]dλ
= −∫ρ(λ)
[A(a, λ)A(b, λ)− A(b, λ)A(b, λ)A(a, λ)A(c , λ)
]dλ
= −∫ρ(λ)A(a, λ)A(b, λ)
[1− A(b, λ)A(c , λ)
]dλ
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 16 / 22
Bell’s inequality
Assuming the probability distribu-tion ρ(λ) is normalized
we note that becauseA(a, λ) = ±1, B(b, λ) = ±1
∫ρ(λ) dλ = 1
P(a, b) ≥ −1
P(a, b) =
∫ρ(λ)A(a, λ)B(b, λ) dλ = −
∫ρ(λ)A(a, λ)A(b, λ) dλ
P(a, b) can only reach -1 when
defining c as a third unit vectora = b
A(a, λ) = −B(a, λ)
P(a, b)− P(a, c) = −∫ρ(λ)
[A(a, λ)A(b, λ)− A(a, λ)A(c , λ)
]dλ
= −∫ρ(λ)
[A(a, λ)A(b, λ)− A(b, λ)A(b, λ)A(a, λ)A(c , λ)
]dλ
= −∫ρ(λ)A(a, λ)A(b, λ)
[1− A(b, λ)A(c , λ)
]dλ
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 16 / 22
Bell’s inequality
P(a, b)− P(a, c) =
∫ρ(λ){−A(a, λ)A(b λ)}
[1− A(b, λ)A(c , λ)
]dλ
The quantity in the square brackets is always either 0 or 2 and the rest,being an integral over a probability density will vary between positive andnegative. Thus, by taking the absolute value of this quantity we can writean overestimate of the quantity by setting the variable part to 1.
|P(a, b)− P(a, c)| ≤∫ρ(λ)
[1− A(b, λ)A(c , λ)
]dλ
≤∫ρ(λ) dλ−
∫ρ(λ)A(b, λ)A(c, λ) dλ
|P(a, b)− P(a, c)| ≤ 1 + P(b, c)
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 17 / 22
Bell’s inequality
P(a, b)− P(a, c) =
∫ρ(λ){−A(a, λ)A(b λ)}
[1− A(b, λ)A(c , λ)
]dλ
The quantity in the square brackets is always either 0 or 2
and the rest,being an integral over a probability density will vary between positive andnegative. Thus, by taking the absolute value of this quantity we can writean overestimate of the quantity by setting the variable part to 1.
|P(a, b)− P(a, c)| ≤∫ρ(λ)
[1− A(b, λ)A(c , λ)
]dλ
≤∫ρ(λ) dλ−
∫ρ(λ)A(b, λ)A(c, λ) dλ
|P(a, b)− P(a, c)| ≤ 1 + P(b, c)
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 17 / 22
Bell’s inequality
P(a, b)− P(a, c) =
∫ρ(λ){−A(a, λ)A(b λ)}
[1− A(b, λ)A(c , λ)
]dλ
The quantity in the square brackets is always either 0 or 2 and the rest,being an integral over a probability density will vary between positive andnegative.
Thus, by taking the absolute value of this quantity we can writean overestimate of the quantity by setting the variable part to 1.
|P(a, b)− P(a, c)| ≤∫ρ(λ)
[1− A(b, λ)A(c , λ)
]dλ
≤∫ρ(λ) dλ−
∫ρ(λ)A(b, λ)A(c, λ) dλ
|P(a, b)− P(a, c)| ≤ 1 + P(b, c)
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 17 / 22
Bell’s inequality
P(a, b)− P(a, c) =
∫ρ(λ){−A(a, λ)A(b λ)}
[1− A(b, λ)A(c , λ)
]dλ
The quantity in the square brackets is always either 0 or 2 and the rest,being an integral over a probability density will vary between positive andnegative. Thus, by taking the absolute value of this quantity we can write
an overestimate of the quantity by setting the variable part to 1.
|P(a, b)− P(a, c)| ≤∫ρ(λ)
[1− A(b, λ)A(c , λ)
]dλ
≤∫ρ(λ) dλ−
∫ρ(λ)A(b, λ)A(c, λ) dλ
|P(a, b)− P(a, c)| ≤ 1 + P(b, c)
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 17 / 22
Bell’s inequality
P(a, b)− P(a, c) =
∫ρ(λ){−A(a, λ)A(b λ)}
[1− A(b, λ)A(c , λ)
]dλ
The quantity in the square brackets is always either 0 or 2 and the rest,being an integral over a probability density will vary between positive andnegative. Thus, by taking the absolute value of this quantity we can write
an overestimate of the quantity by setting the variable part to 1.
|P(a, b)− P(a, c)| ≤∫ρ(λ)
[1− A(b, λ)A(c , λ)
]dλ
≤∫ρ(λ) dλ−
∫ρ(λ)A(b, λ)A(c, λ) dλ
|P(a, b)− P(a, c)| ≤ 1 + P(b, c)
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 17 / 22
Bell’s inequality
P(a, b)− P(a, c) =
∫ρ(λ){−A(a, λ)A(b λ)}
[1− A(b, λ)A(c , λ)
]dλ
The quantity in the square brackets is always either 0 or 2 and the rest,being an integral over a probability density will vary between positive andnegative. Thus, by taking the absolute value of this quantity we can writean overestimate of the quantity by setting the variable part to 1.
|P(a, b)− P(a, c)| ≤∫ρ(λ)
[1− A(b, λ)A(c , λ)
]dλ
≤∫ρ(λ) dλ−
∫ρ(λ)A(b, λ)A(c, λ) dλ
|P(a, b)− P(a, c)| ≤ 1 + P(b, c)
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 17 / 22
Bell’s inequality
P(a, b)− P(a, c) =
∫ρ(λ){−A(a, λ)A(b λ)}
[1− A(b, λ)A(c , λ)
]dλ
The quantity in the square brackets is always either 0 or 2 and the rest,being an integral over a probability density will vary between positive andnegative. Thus, by taking the absolute value of this quantity we can writean overestimate of the quantity by setting the variable part to 1.
|P(a, b)− P(a, c)| ≤∫ρ(λ)
[1− A(b, λ)A(c , λ)
]dλ
≤∫ρ(λ) dλ−
∫ρ(λ)A(b, λ)A(c, λ) dλ
|P(a, b)− P(a, c)| ≤ 1 + P(b, c)
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 17 / 22
Bell’s inequality
P(a, b)− P(a, c) =
∫ρ(λ){−A(a, λ)A(b λ)}
[1− A(b, λ)A(c , λ)
]dλ
The quantity in the square brackets is always either 0 or 2 and the rest,being an integral over a probability density will vary between positive andnegative. Thus, by taking the absolute value of this quantity we can writean overestimate of the quantity by setting the variable part to 1.
|P(a, b)− P(a, c)| ≤∫ρ(λ)
[1− A(b, λ)A(c , λ)
]dλ
≤∫ρ(λ) dλ−
∫ρ(λ)A(b, λ)A(c, λ) dλ
|P(a, b)− P(a, c)| ≤ 1 + P(b, c)
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 17 / 22
Bell’s inequality
Take the particular case of a ⊥ b
and c at an angle γ to a.
P(a, b) = 0
P(a, c) = − cos γ
P(b, c) = − cos(π
2− γ)
= − sin γ
Bell’s inequality becomes
| cos γ| ≤ 1− sin γ
a
b
cγ
Clearly this inequality does not hold for all cases meaning that anexperimental result predicted by quantum mechanics woud rule out anyhidden variables
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 18 / 22
Bell’s inequality
Take the particular case of a ⊥ b
and c at an angle γ to a.
P(a, b) = 0
P(a, c) = − cos γ
P(b, c) = − cos(π
2− γ)
= − sin γ
Bell’s inequality becomes
| cos γ| ≤ 1− sin γ
a
b
cγ
Clearly this inequality does not hold for all cases meaning that anexperimental result predicted by quantum mechanics woud rule out anyhidden variables
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 18 / 22
Bell’s inequality
Take the particular case of a ⊥ band c at an angle γ to a.
P(a, b) = 0
P(a, c) = − cos γ
P(b, c) = − cos(π
2− γ)
= − sin γ
Bell’s inequality becomes
| cos γ| ≤ 1− sin γ
a
b
cγ
Clearly this inequality does not hold for all cases meaning that anexperimental result predicted by quantum mechanics woud rule out anyhidden variables
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 18 / 22
Bell’s inequality
Take the particular case of a ⊥ band c at an angle γ to a.
P(a, b) = 0
P(a, c) = − cos γ
P(b, c) = − cos(π
2− γ)
= − sin γ
Bell’s inequality becomes
| cos γ| ≤ 1− sin γ
a
b
cγ
Clearly this inequality does not hold for all cases meaning that anexperimental result predicted by quantum mechanics woud rule out anyhidden variables
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 18 / 22
Bell’s inequality
Take the particular case of a ⊥ band c at an angle γ to a.
P(a, b) = 0
P(a, c) = − cos γ
P(b, c) = − cos(π
2− γ)
= − sin γ
Bell’s inequality becomes
| cos γ| ≤ 1− sin γ
a
b
cγ
Clearly this inequality does not hold for all cases meaning that anexperimental result predicted by quantum mechanics woud rule out anyhidden variables
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 18 / 22
Bell’s inequality
Take the particular case of a ⊥ band c at an angle γ to a.
P(a, b) = 0
P(a, c) = − cos γ
P(b, c) = − cos(π
2− γ)
= − sin γ
Bell’s inequality becomes
| cos γ| ≤ 1− sin γ
a
b
cγ
Clearly this inequality does not hold for all cases meaning that anexperimental result predicted by quantum mechanics woud rule out anyhidden variables
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 18 / 22
Bell’s inequality
Take the particular case of a ⊥ band c at an angle γ to a.
P(a, b) = 0
P(a, c) = − cos γ
P(b, c) = − cos(π
2− γ)
= − sin γ
Bell’s inequality becomes
| cos γ| ≤ 1− sin γ
a
b
cγ
Clearly this inequality does not hold for all cases meaning that anexperimental result predicted by quantum mechanics woud rule out anyhidden variables
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 18 / 22
Bell’s inequality
Take the particular case of a ⊥ band c at an angle γ to a.
P(a, b) = 0
P(a, c) = − cos γ
P(b, c) = − cos(π
2− γ)
= − sin γ
Bell’s inequality becomes
| cos γ| ≤ 1− sin γ
a
b
cγ
Clearly this inequality does not hold for all cases meaning that anexperimental result predicted by quantum mechanics woud rule out anyhidden variables
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 18 / 22
Bell’s inequality
Take the particular case of a ⊥ band c at an angle γ to a.
P(a, b) = 0
P(a, c) = − cos γ
P(b, c) = − cos(π
2− γ)
= − sin γ
Bell’s inequality becomes
| cos γ| ≤ 1− sin γ 0
0.5
1
1.5
2
0 π 2π
γ
a ⊥ b
|cos γ| 1-sin γ
Clearly this inequality does not hold for all cases meaning that anexperimental result predicted by quantum mechanics woud rule out anyhidden variables
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 18 / 22
Bell’s inequality
Take the particular case of a ⊥ band c at an angle γ to a.
P(a, b) = 0
P(a, c) = − cos γ
P(b, c) = − cos(π
2− γ)
= − sin γ
Bell’s inequality becomes
| cos γ| ≤ 1− sin γ 0
0.5
1
1.5
2
0 π 2π
γ
a ⊥ b
|cos γ| 1-sin γ
Clearly this inequality does not hold for all cases
meaning that anexperimental result predicted by quantum mechanics woud rule out anyhidden variables
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 18 / 22
Bell’s inequality
Take the particular case of a ⊥ band c at an angle γ to a.
P(a, b) = 0
P(a, c) = − cos γ
P(b, c) = − cos(π
2− γ)
= − sin γ
Bell’s inequality becomes
| cos γ| ≤ 1− sin γ 0
0.5
1
1.5
2
0 π 2π
γ
a ⊥ b
|cos γ| 1-sin γ
Clearly this inequality does not hold for all cases meaning that anexperimental result predicted by quantum mechanics woud rule out anyhidden variables
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 18 / 22
The EPR experiment
“Experimental test of Bell’s inequalities usint time-varying analyzers,”, A. Aspect, J.
Dalibard, and G. Roger, Phys. Rev. Lett. 49, 1804-1807 (1982).
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 19 / 22
The EPR experiment
“Experimental test of Bell’s inequalities usint time-varying analyzers,”, A. Aspect, J.
Dalibard, and G. Roger, Phys. Rev. Lett. 49, 1804-1807 (1982).
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 20 / 22
The EPR experiment
“Experimental test of Bell’s inequalities usint time-varying analyzers,”, A. Aspect, J.
Dalibard, and G. Roger, Phys. Rev. Lett. 49, 1804-1807 (1982).
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 21 / 22
The EPR experiment
“Experimental test of Bell’s inequalities usint time-varying analyzers,”, A. Aspect, J.
Dalibard, and G. Roger, Phys. Rev. Lett. 49, 1804-1807 (1982).
C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 22 / 22